A treatise of angular sections by John Wallis ...
Wallis, John, 1616-1703., Wallis, John, 1616-1703. Treatise of algebra.
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A TREATISE OF Angular Sections.

By JOHN WALLIS, D. D. Professor of Geometry in the University of Oxford, and a Member of the Royal Society, LONDON.

[illustration] [printer's or publisher's device]

LONDON: Printed by Iohn Playford, for Richard Davis, Bookseller, in the University of OXFORD, 1684.

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A TREATISE OF Angular Sections.

CHAP. 1. Of the Duplication and Bisection of an ARCH or ANGLE.

I. LET the Chord (or Subtense) of an Arch proposed, be called A, (or E;) of the Double, B; of the Treble, C; of the Quadruple, D; of the Quintruple, F; &c. The Radius, R; the Diameter, 2R. (But sometimes we shall give the name of the Subtense A, E, &c. to the Arch whose Subtense it is; yet with that care, as not to be liable to a mistake.)

II. Where the Subtense of an Arch is A; let the Versed sine be V: (where* that is E, let this be U.) Which drawn into (or Multiplied by) the remainder of the Diameter (2 R − V) makes 2 R VVq, the Square of the Right-sine: (this Sine being a Mean-proportional between the Segments of the Diameter on which it stands erect, by 13 6.) That is, (Q: ½ B:) the Square of (the Right∣sine, or) half the Subtense of the double Arch: That is, 2 R V−Vq = Q: ½ B: = ¼ Bq.

III. If to this we add Vq (the Square of the Versed-sine,) it makes 2 RV = (¼ Bq + Vq =) Aq. (And, by the same reason, 2 R U=Eq.) That is,

IV. The Subtense of an Arch, is a Mean Proportional between the Diameter and the Versed-sine.

V. Again, because 2 R V=Aq, therefore (dividing both by 2 R,) 〈 math 〉: And (the Square thereof) 〈 math 〉 Which subtracted from Aq, leaves the Square of the Right-sine, 〈 math 〉 (And, in like manner, 〈 math 〉, and 〈 math 〉, and 〈 math 〉 That is,

Page  2VI. If from the Square of the Subtense, we take its Biquadrate divided by the Square of the Diameter; the Remainder is equal to the Square of the Right-sine: And the Square-root of that Remainder, to the Sine it self: And, the double of this, to the Subtense of the double Arch.

VII. Accordingly, because 〈 math 〉 therefore (its Quadruple) 〈 math 〉 and 〈 math 〉 (And in like manner, 〈 math 〉 That is,

VIII. If from Four-times the Square of a Subtense, are taken its Biquadrate divided by the Square of the Radius; the Remainder is the Square of the Subtense of the double Arch: And, the Quadratick Root of that Remainder, is the Subtense it self.

IX. But 〈 math 〉 That is,

X. The Rect-angle of the Subtense of an Arch, and of its Remainder to a Semicircle, divided by the Radius; is equal to the Subtense of the double Arch.

XI. Because 〈 math 〉; therefore AE = RB = 2R × ½B: And, R. A :: B. B: And therefore, (because A E contain a Right-angle, as being an Angle in a Semicircle.)

XII. In a Right-angled Triangle, the Rect-angle of the two Legs containing the Right-angle, is equal to that of the Hypothenuse, and the Perpendicular from the Right-angle thereupon. And,

XIII. As the Radius, to the Subtense of an Arch; so the Subtense of its Remainder to a Semicircle, is to that of the double Arch.

XIV. Because B, the Subtense of a double Arch, doth indifferently subtend the two Segments which compleat the whole Circumference; and, consequently, the half of either may be the single Arch of this double: It is therefore necessary that this Equation have two (Affirmative) Roots; the greater of which we will call A; and the lesser E: And therefore 〈 math 〉 That is,

XV. Any Arch, and its Remainder to a Semicircumference, (as also its excess above a Semicircumference, and either of them increased by one or more Semicircum∣ferences,) will have the same Subtense of the double Arch. For in all these Cases, the Subtense of the single Arch will be either A or E.

XVI. Because 〈 math 〉 And therefore, 〈 math 〉 and 4Aq Rq − Aqq (= Bq Rq) = 4Eq Rq − Eqq: Therefore (by Transposition) 4Aq Rq − 4Eq Rq = Aqq − Eqq; and (dividing both by 〈 math 〉 That is,

Page  3XVII. The Square of the Diameter, is equal to the difference of the Biquadrates of the Subtenses of two Arches, (which together complete a Semicircumference) divided by the difference of their Squares: And this also, equal to the sum of the Squares of those Subtenses. That is, (because A E contain a Right-angle.)

XVIII. In a Right-angled Triangle, the Square of the Hypothenuse (4Rq) is equal to the Squares of the sides containing the Right-angle. (Aq+Eq.)

XIX. Or thus, Because B is the common Subtense to two Segments, which* together complete the whole Circumference; and therefore the half of both, complete the Semicircumference: If therefore in a Circle (according to Ptolemy's Lemma) a Trapezium be inscribed, whose opposite sides are A, A; and E, E: The Diagonals will be Diameters, that is, 2R: And, consequently, 4Rq = Aq + Eq; as before.

XX. Hence therefore, The Radius (R) with the Subtense of an Arch (A or E) being given; we have thence the Subtense of the double Arch, B: (which is the Duplication of an Arch or Angle.) For, R, A, being given; we have 〈 math 〉 (or R, E, being given, we have A = 4Rq − Eq:) And, having R, A, E; we have 〈 math 〉, by § 9.

XXI. The Radius R, with B the Subtense of the double Arch, being given; we have thence the Subtense of the single Arch, A or E. (which is the Bisection of an Arch or Angle.) For, by § 14, 〈 math 〉: And therefore 4Rq Aq − Aqq (= Rq Bq) = 4Rq Eq − Eqq. And the Roots of this Equation, 〈 math 〉, or Eq. And, the Quadratick Root of this, is A, or E.

XXII. Hence also we have an easie Method, for a Geometrical Construction for*the Resolution of such Biquadratick Equations; or Quadratick Equations of a Plain Root, wherein the Highest Power is Negative. (Understand it in Mr. Oughtred's Language: Who puts the Absolute Quantity, Affirmative; and by it self; and the rest of the Equation all on the other side.) Suppose, Rq Bq = 4Rq Aq − Aqq, or (putting P = ½ B,) 4Rq Pq = 4Rq Aq − Aqq. For, dividing the Absolute term Rq Bq, or 4Rq Pq, by the Co-efficient of the middle term 4Rq, the Result is ¼Bq, or Pq; and its Root ½ B or P. Which being set Perpendicular on a Diameter equal to 2 R (the Square Root of that Co-efficient:) a streight Line from the top of it, Parallel to that Diameter, will (if the Equation be not impossible) cut the Circle, or at least touch it: From which Point of Section or Contact, two streight-lines drawn to the ends of the Diameter, are A, and E, the two Roots of that (ambiguous) Biquadratick Equation, (or, if we call it a Quadratick of a Plain-root, the Root of the Plain-root of such Quadratick Equation.)

XXIII. And this Construction, is the same with the Resolution of this Problem; In a Right-angled Triangle, the Hypothenuse being given, and a Perpendicular from the Right-angle thereupon, to find the other sides; (and, if need be, the Angles, the Segments of the Hypothenuse, and the Area of the Triangle ½ R B or P R.)

XXIV. Or thus: Having R and B, (as at § 22.) with the Radius R describe a Circle; and therein inscribe the Chord B; and another on the middle hereof at Right-angles: (which will therefore bisect that, and be a Diameter:) And, from both ends of this, to either end of B, draw the Lines A, E; as before. And this Construction is better than the former, because of the uncertainty of the precise Point of Contact or Section, in case the Section be somewhat Oblique.

Page  4XXV. Now if it be desired, in like manner, to give a like Construction, in Case of such Biquadratick Equations (or Quadraticks of a Plain-root) where the highest Power is Affirmative; (though that be here a Digression, as in all the rest that follow, to § 35.) It is thus: Suppose the Equations Aqq − VqAq = VqEq = Pqq + VqPq: Whose (Affirmative) Roots are Aq, and Pq; (and therefore Vq, VqEq, and consequently Eq, are known Quantities:) Therefore (by Transposition) Aqq − VqAq + VqPq; and (dividing by 〈 math 〉 And therefore Aq − Vq = Pq; and Pq + Vq = Aq: And (by Multiplication) Aqq − VqAq = AqPq = Pqq + VqPq = VqEq.

〈 math 〉

XXVI. The Equation therefore proposed (dividing all by Vq) comes to this, 〈 math 〉 That is, 〈 math 〉. Whose Roots are 〈 math 〉, and 〈 math 〉. Namely, 〈 math 〉. And 〈 math 〉. And these Multiplied into V (a known Quantity) make Aq, and Pq: Namely, 〈 math 〉. And 〈 math 〉 And consequently, A is a mean Proportional between V and 〈 math 〉. And P, a mean Proportional between V and 〈 math 〉. Therefore,

XXVII. And Equation being proposed in one of these Forms, Aqq − VqAq* = VqEq = Pqq + VqPq: The absolute term (VqEq) being divided by the Co-efficient of the middle term (Vq;) the quantity resulting is (Eq) whose Square-root (E,) set Perpendicular on the end of a streight Line equal to (V) the Square-root of the Co-efficient; which we may suppose the Diameter of a Circle, to which that Perpendicular is a Tangent: On the same Center with this Circle (and on the same Diameter continued) by the Top of that Perpendicular, draw a second Circle. The Diameter of this second Circle, is, by that Perpendi∣cular (E) cut into two Segments, which are the Roots of these Equations. That is, 〈 math 〉; and 〈 math 〉

XXVIII. Or, (without drawing that second Circle,) from the Top of that Perpendicular (in a streight Line through the Center of the first, which will cut the Circumference in two Points,) to the first Section, is 〈 math 〉; to the second, 〈 math 〉

XXIX. These two Roots, Multiplied one into the other, become equal to the Absolute quantity, 〈 math 〉. And Multiplied into V, become Aq, Pq: Or, thus, P is a mean Proportional between V and U; and A, between V and V+U: Or thus, P is a mean Proportional between V and 〈 math 〉; and (because by § 25. Aq = Pq + Vq,) A is the Hypothenuse (in a Right-angled Triangle) to the Legs P, V. And this is no contemptible Method, For the resolving Quadra∣tick Equations of a Plain-root, wherein the highest term is Affirmative. The whole Geometrick Construction, is clear enough from the Figures adjoined; where yet the Circles, (for the most part) serve rather for the Demonstration, than the Construction.

Page  5XXX. Again, (by the same § 25.) 〈 math 〉 And there∣fore A and E, are also the Legs of a Right-angled Triangle, whose Hypothe∣nuse is V + U: Which, by P (a Perpendicular on it from the Right-angle) is cut into those two Segments.

XXXI. From the same Construction therefore, we have also the Geometrical Construction of this Problem; In a Right-angled Triangle, having one of the Legs E, with the farther Segment of the Hypothenuse V, to find the other Segment; (and so the whole V + U; and the Perpendicular P; and the other Leg A; and the whose Triangle.)

XXXII. We have thence also this Analogy; 〈 math 〉 And 〈 math 〉 Or thus, 〈 math 〉 And 〈 math 〉

XXXIII. If therefore we make V the Radius of a Circle; then is A the* Secant; P, the Tangent; E a Parallel to the Right-sine (in contrary position) from the end of the Secant to the Diameter produced. If we make A the Ra∣dius; then is P the Right-sine, and E the Tangent of the same Arch; and V, the Sine of the Complement, or Difference between the Radius and Versed Sine. From hence therefore,

XXXIV. The Tangent E, and Sine of the Complement V, being given; we have the Right-sine P, and the Radius A. (But, § 25, and all hitherto, is a Digression.)

XXXV. If in a Semicircle on the Diameter 2 R, we inscribe B the Subtense* of a double Arch: A Perpendicular on the middle Point hereof, will cut the Arch of that Semicircle into two Segments, (whose Subtenses are A, E;) either of which is a single Arch, to the double whereof, B is a Subtense. This, as to E, is evident from 4 è 1, and 28 è 3: And, as to A, from § 15 of this.

XXXVI. But also (by the same reason,) the Arch β (the difference of the Arches A, E;) and B (the double of either,) will (if doubled) have the same Subtense of their double Arch. That is, The double of the double (of either) and the double of their difference, will have the same Subtense.

XXXVII. If an Arch to be doubled, be just a third part of the Circumference;* the Subtense of the double, is equal to that of the single Arch. (For the same Subtense, which on one side subtends two Trients, doth on the other side subtend but one.) That is, by § 7, 〈 math 〉. And there∣fore (by Transposition) 〈 math 〉, and 3Rq = Aq. That is,

XXXVIII. The Square of the Subtense to a Trient of the Circumference (or of the side of an Equilater Triangle inscribed) is equal to three Squares of the Radius.

XXXIX. Again, the same being the Subtense of the double Trient, and of the double Sextant, (for a Trient and a Sextant compleat the half, ⅓ + ⅙ = ½) the Square of the Subtense of a Sextant, (Eq,) is the difference of the Squares of that of the Trient, and (the Diameter or) that of the Semicircumference. That is, 4Rq − Aq = Eq; that is, (by § preced.) 4Rq − 3Rq = Rq = Eq: And, E = R. That is,

XL. The Subtense of a Sextant (or side of the inscribed Equilater Hexagon) is equal to the Radius.

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CHAP. II. Of the Triplication and Trisection of an ARCH or ANGLE.

I. IF in a Circle, be inscribed a Quadrilater, whose three sides are A, A, A,* (Subtenses of a single Arch) and the fourth C, (the Subtense of the Triple Arch:) the Diagonals are B, B, the Subtense of the double;) as is evident. But it is evident also, that (in this Case) A is less than a Trient of the whole Circumference.

II. And therefore (the Rect-angle of the Diagonals being equal to the two Rect-angles of the opposite sides,) Bq = Aq+AC; and therefore Bq−Aq = AC, and 〈 math 〉. That is,

III. The Square of the Subtense of the double Arch; is equal to the Square of the Subtense of the Single Arch (less than a Trient of the Circumference) and the Rect∣angle of the Subtenses of the Sngle and Treble Arch. And therefore,

IV. The Square of the Subtense of the double Arch, wanting the Square of the Subtense of the single Arch, (being less than a Trient,) is equal to the Rect-angle of the Subtenses of the single and Treble Arch. And consequently,

V. If the Square of the Subtense of the double Arch, wanting the Square of the Subtense of the single Arch (less than a Trient,) be divided by the Subtense of the single Arch; the Result is the Subtense of the Triple Arch.

VI. Because that (by § 2.) 〈 math 〉; and, that B+A into B−A is equal to Bq−Aq: (as will appear by Multiplication:). Therefore, 〈 math 〉 That is,

VII. As the Subtense of a single Arch (less than a Trient) to the sum of the Subtenses of the single and double Arch; so is the Excess of that of the double above that of the single, to the Subtense of the Triple.

VIII. Again, because (by § 7, of the precedent Chapter,) 〈 math 〉: Therefore, 〈 math 〉: And therefore 〈 math 〉. That is,

IX. The Triple of the Subtense of an Arch (less than a Trient,) wanting the Cube thereof, divided by the Square of the Radius; is equal to the Subtense of the Triple Arch.

X. But, because the same Subtense C, subtends also to another Segment of the same Circle; the Subtense of whose Trient we shall call E: Therefore 〈 math 〉.

XI. And because the three Arches A, A, A; and the three Arches E, E, E; complete the whole Circumference: (as is evident;) Therefore, once A, and once E, complete a Trient or third part thereof. Therefore,

Page  7XII. An Arch less than the Trient of a Circumference, and the Residue of that*Trient, (A, and E,) have the same Subtense of their Triple Arch.

XIII. Again, because (as is shewed already) 〈 math 〉; and therefore 3RqA−Ac = 3RqE−Ec; and 3RqA−3RqE = Ac−Ec: Therefore, (dividing both by A−E,) 〈 math 〉. (As will appear upon dividing Ac−Ec by A−E; or Multiplying A−E into Aq+AE+Eq.)

XIV. But (by § 37, 38, Chap. preced.) 3Rq is the Square of the Subtense of a Trient; that is (by § 11 of this) of the sum of the Arches A and E. There∣fore,

XV. The Square of the Subtense of the Trient of the Circumference of a Circle, (or three Squares of the Radius,) is equal to the Squares of the Subtenses of any two Arches completing that Trient, and the Rect-angle of them. That is, (putting T for the Subtense of a Trient) Tq (= 3Rq) = Aq+AE+Aq.

XVI. But the Angle which AE contain, (as being an Angle in the Trient of a Circle, or insisting on two Trients,) is an Angle of 120 Degrees. And there∣fore (by § 15.)

XVII. In a Right-lined Triangle, one of whose Angles is 120 Degrees; the Square of the Subtense to that Angle, is equal to the two Squares of the sides containing it, and a Rect-angle of those sides. (For, if such Triangle be inscribed in a Circle, the Base of that Triangle, will be the subtendent of a Trient in such Circle; or 〈 math 〉 Rq.)

XVIII. If a Quadrilater be inscribed in a Circle, three of whose sides are* A, E, A, (or E, A, E,) and the fourth Z: Each of the Diagonals (by § 11.) is T, the Subtense of a Trient. And therefore (by § 13, 14, 15,) ZE+Aq (= ZA+Eq) = Tq = 3Rq = Aq+AE+Eq. And, consequently, ZE = AE+Eq, and ZA = Aq+AE; and therefore, Z = A+E. And therefore,

XIX. If to the Aggregate of two Arches A, E, (completing a Trient,) be added a third equal to either of them; Z, the Subtense of the Aggregate of all the three, is equal to the sum of the Subtenses of those two. That is, Z = A+E.

XX. But the same Chord Z, doth subtend, on the one side, to a Trient increased by the Arch A; and, on the other side, to a Trient increased by the Arch E; (as is evident;) That is, to an Arch which doth as much exceed a Trient (or want of two Trients,) as the Arch A or E wants of a Trient. Therefore,

XXI. The Aggregate of the Subtenses of two Arches, which together make up a Trient; is equal to the Subtense of another Arch which doth as much exceed a Trient, (or want of two Trients,) as either of those two wants of a Trient.

XXII. The same will in like manner be inferred, if we inscribe a Quadrilater* whose opposite sides are A, T, and E, T; and the Diagonals TZ. For then TA+TE = TZ; and therefore A+E = Z, as before.

Page  8XXIII. But if either of the Arches to which Z subtendeth (greater than a* Trient, and less than two Trients) be Tripled; the Subtense of this Triple, is the same with that of the Triple of A or E. For the Triple of an Arch greater than a Trient, is equal to one whole Circumference with the Triple of that Excess. (For the Triple of ⅓+A, is 1+3A.) Now because, when we have once gone round the whole Circumference, we are just there where at first we began; this therefore (as to this Point) is as nothing; and the whole distance to be acquired is but the Triple of such Excess; and just the same as if onely this Excess had been thrice taken.

XXIV. As for Example: If the Arch subtended by Z, be β γ δ, (that is, a* Trient increased by the Arch E;) and to this we add a second equal to it, δ ζ θ; the Aggregate β γ δ ζ θ, is the double Arch, and the Subtense thereof is B, or β θ, (which is also the Subtense of the Difference of the Arches A, E:) and if to these two, we add a third equal to either of them θ γ χ; then is β γ δ ζ θ γ χ, the Triple of the Arch first proposed; and the Subtense hereof (that is, the Streight-line which joyns the beginning and the end of this Triple Arch) is β χ = C; the very same which subtends the Triple of E.

XXV. And just the same would come to pass, if for the first Arch we take β θ ζ δ (that is, a Trient increased by A,) to which Z is a subtendent likewise. For, taking a second equal to it δ χ γ β θ; the Aggregate β θ ζ δ χ γ β θ (more than one entire Circumference) is the double Arch, and the Subtense thereof B as before: And if to these two we add a third equal to either, θ ζ δ χ; the Triple Arch is β θ ζ δ χ γ β θ ζ δ χ; and the Subtense hereof (as before) β χ or C; the same with the subtendent of the Triple of A. And therefore,

XXVI. The Triple of an Arch greater than a Trient, hath the same Subtense with the Triple of its Excess above a Trient. And the same (for the same reason) holds in Arches greater than 2, 3, or more Trients.

XXVII. But note here, that, in this case; That is, if the Arch to be Tripled be greater than a Trient, but less than two Trients, (for if more than two Trients, but less than the whole Circumference, it is the same as if it were less than a Tri•••••;) the Subtense of the double is less than that of the single. For, in such case, the Arch will differ from that of a Semicircle (either in Excess or in Defect) by less than ⅙ of the whole Circumference. Let it be X. If there∣fore ½ ± X be the single Arch, the double will be 1 ± 2X; and the Subtense thereof (whether greater or less than one entire revolution) will be the same with that of 2X: And therefore (X being less than ⅙,) 2X will require a less Subtense than that of ½ ± X; that being less than the Subtense of a Trient, but this greater than it. And the like is to be understood in other cases of like nature.

XXVIII. Supposing therefore, as before, 〈 math 〉, or 〈 math 〉; C must in this case be a Negative quantity: Or, if we put C Affirmative, then must Z be Negative, (or less than nothing:) For Bq−Zq (where a greater quantity is to be subtended from a less) must needs be Negative; that is Bq−Zq = Zc; where Zc being a Negative, either Z or C must be so too, or else (putting all Affirmative) Zq−Bq = Zc, and Zq = Bq+ZC.

XXIX. Which is evident also from the Diagram; where, for this reason, ZZ become Diagonals; and both BB, and ZC, opposite sides. And therefore Zq−Bq = ZC, or Zq = Bq+Zc; and 〈 math 〉. That is,

Page  9XXX. The Square of the Subtense of a single Arch, greater than a Trient, but*less than two Trients; is equal to the Square of the Subtense of the double Arch, together with a Rect-angle of the Subtenses of the single and Triple Arch. And,

XXXI. The Square of the Subtense of a single Arch (greater than a Trient, but less than two Trients,) wanting the Square of the Subtense of the double Arch; is equal to the Rectangle of the Subtenses of the single and Triple Arch. And therefore,

XXXII. If the Square of the Subtense of a single Arch (greater than a Trient, but less than two Trients,) wanting the Square of the Subtense of the double Arch, be divided by that of the single, the Result is the Subtense of the Triple Arch: (Or, if divided by that of the Triple, the Result is that of the single.) Or,

XXXIII. If from the Subtense of a single Arch (greater than a Trient but less than two Trients,) we Subtract the Square of the Subtense of the double Arch divided by that of the single; the Remainder is equal to the Subtense of the Triple.

XXXIV. But, because of 〈 math 〉, or Z) Zq−Bq (C; and Zq−Bq = Z+B into Z−B: We have thence this Analogy, 〈 math 〉 That is,

XXXV. As the Subtense of a single Arch (greater than a Trient but less than two Trients,) to the Aggregate of the Subtenses of the single and double; so is that of the single wanting that of the double to that of the Triple.

XXXVI. Now because (as we have shewed) Zq−Bq = ZC; and (by § 7, 8, Chap. preced.) 〈 math 〉: Therefore 〈 math 〉: And 〈 math 〉. That is,

XXXVII. If from the Cube of the Subtense of a single Arch (greater than a Trient but less than two Trients) divided by the Square of the Radius; we subtract the Triple of that Subtense: The Remainder is equal to the Subtense of the Triple Arch.

XXXVIII. If the Arch to be Tripled be greater than two Trients, it is the same as if it were less than one Trient. (For the Residue of the whole, to which it also subtends, is then less than a Trient.) And therefore the same Chord (suppose A or E) subtends as well to an Arch greater than two Trients, as to one less than one Trient.

XXXIX. If the Arch to be Tripled be equal to a Trient; it is indifferent to whether of the two cases it be referred, (that of the greater, or that of the lesser, than a Trient,) and the same happens if it be supposed equal to two or more Trients, or to one or more intire revolutions.

XL. If the Arch to be Tripled be greater than one or more intire revolutions; its Subtense is the same with that of its Excess above those intire revolutions, and to be considered in like manner, which things are evident and need no further demonstration.

XLI. Now what hath been severally delivered concerning the Triplication of an Arch or Angle less than a Trient; and of one greater than a Trient, but less than two Trients; (to one of which cases every Arch may be referred, as is already shewed;) we may thus, jointly put together.

Page  10XLII. The Difference of the Squares of the Subtenses of the single and double Arch*(whether soever of them be the greater,) is equal to the Rect-angle of the Subtenses of the single and Triple. (by § 4 and 31.) That is, Bq − Aq = AC, and Zq − Bq = ZC. And therefore,

XLIII. If the Difference of the Squares of the Subtenses of the single and double Arch, be divided by the Subtense of the single; it gives that of the Triple: If, by that of the Triple, it gives that of the single. (by § 5, 32.) That is, 〈 math 〉. And 〈 math 〉. And 〈 math 〉. And,

XLIV. As the Subtense of the single Arch, to the Aggregate of the Subtenses of the single and double; so is the Difference of these Subtenses, to the Subtense of the Triple. (by § 7, 35.) That is, 〈 math 〉 And 〈 math 〉

XLV. Now for as much as (by § 12, 26.) the Three Arches A, E, Z, if Tripled, will have the same Subtense of the Triple Arch C: 'Tis thence manifest, that such Equation as this (which concerns the Trisection of an Arch,)〈 math 〉must have in all Three Roots, as A, E, Z: (For every of these, upon such Triplication, will have the Subtense of the Triple Arch, C:) Yet so; that, where A and E are Affirmative Roots, Z is a Negative; and contrarywise, where this is Affirmative, those be Negative. That is, in this Equation, 〈 math 〉; the Roots are + A, + E − Z. But, in this, 〈 math 〉; the Roots are − A, − E, + Z. And therefore, 〈 math 〉. And, consequently, 3ARq − Ac = 3ERq − Ec = CRq = Zc − 3ZRq.

XLVI. Since therefore 3ARq − Ac = Zc − 3ZRq; and consequently (by Transposition) 3ZRq + 3ARq = Zc + Ac; it is also (dividing both sides by Z + A,) 〈 math 〉. (For Z + A into Zq − ZA + Aq, is equal to Zc + Ac; as will appear by Multiplication; and contrarywise, if this be divided by either of those, the Quotient will be the other of them, as will be found by Division.) And in like manner; because also 3ERq − Ec = Zc − 3ZRq, therefore 3ZRq + 3ERq = Zc + Ec; and 〈 math 〉. That is,

XLVII. Of two Arches, whereof the one exceeds the other by a Trient of the whole Circumference; or else, whereof the one doth as much exceed a Trient as the other wants of it; the Squares of the Subtenses, wanting a Rect-angle of the same Subtenses, are equal to the Square of the Subtense of a Trient, or Three times the Square of the Radius. That is, Zq − ZA + Aq = 3Rq = Tq = Zq − ZE + Eq.

XLVIII. Now the Angle contained by the Legs ZA, or ZE, (standing on the Chord T,) is an Angle of 60 Degrees; (as being an Angle in the Circum∣ference standing on an Arch of 120 Degrees.) And therefore,

XLIX. In a Right-lined Triangle, one of whose Angles is of 60 Degrees; the*Square of the side opposite to this Angle, is equal to the two Squares of the sides contain∣ing it, wanting the Rect-angle of the same sides. (For any such Triangle may be thus inscribed in a Circle:) That is, Zq − ZA + Aq, (or Zq − ZE + Eq,) = Tq = 3Rq.

Page  11L. The same things (from § 45, &c.) may be thus otherwise inferred.* Because (by § 15) Aq + AE + Eq = 3Rq, and (by § 18 or 21.) Z = A + E; therefore Zq = Aq + 2AE + Eq, and ZA = Aq + AE, (and ZE = AE + Eq,) and therefore Zq − ZA = AE + Eq, (and Zq − ZE = Aq + AE;) and consequently Zq − ZA + Aq (or Zq − ZE + Eq,) = Aq + AE + Eq = 3Rq. From whence the rest are inferred as before.

〈 math 〉

LI. Moreover, because (as is before shewed) 〈 math 〉: We may thence infer the following Theorems.

LII. The Difference of the Cubes of two Legs containing an Angle of 120 degrees, divided by the Difference of those Legs, is equal to the Square of the Base subten∣ded to it.

LIII. But if it be an Angle of 60 degrees; the sum of the Cubes of the Legs or sides containing it, divided by the sum of those sides, is equal to the Square of the Base. Again,

LIV. The Difference of the Legs containing an Angle of 120 degrees, Multiplied into the Square of the Base, is equal to the Difference of the Cubes of those Legs.

LV. But if it be an Angle of 60 degrees; the sum of the Legs Multiplied into the Square of the Base is equal to the sum of the Cubes of those Legs.

LVI. Again, because 〈 math 〉 (or 〈 math 〉 and 〈 math 〉: Therefore 〈 math 〉, (and 〈 math 〉,) and 〈 math 〉. And therefore,

LVII. The Difference between the Triple of the Subtense of a single Arch, less than a Trient; and, of the Subtense of the Triple of that Arch; is equal to the Cube of the Subtense of that single Arch, divided by the Square of the Radius. And consequently, That Difference Multiplied into the Square of the Radius, is equal to such Cube.

LVIII. The sum of the Triple of the Subtense of a single Arch greater than a Trient, but less than two Trients, and of the Subtense of the Triple Arch; is equal to the Cube of the Subtense of that single Arch divided by the Square of the Radius. And consequently, That sum Multiplied into the Square of the Radius, is equal to such Cube.

LIX. Because (as is before shewed) 〈 math 〉: Or, 3ARq − AC = 3ERq − Ec = CRq = Zc − 3ZRq: There∣fore the Subtense of an Arch being given (as A, E, or Z,) together with the Radius R; we have thence the Subtense of the Triple Arch C. Which is, the Triplication of an Arch or Angle.

Page  12LX. And, contrarywise; The Radius of a Circle R, and the Subtense of the Triple*Arch C, being given; we have thence the Subtense of the single Arch, (A, E, or Z,) by resolving such a Cubick Equation. Which is, the Trisection of an Arch or Angle. But the Geometrical effection thereof is not to be performed by Rule and Com∣pass; without the help of a Conick Section, or some Line more Compounded.

LXI. But then, on the other side; such Cubick Equations may be resolved by the Trisection of an Arch. For, suppose a Cubick Equation of this Form, 3RqA − Ac, (or 3RqE − Ec,) = RqC; whose Root A (or E) is sought. Now if R (the Square-root of a third part of the Co-efficient) be made the Radius of a Circle, (that is, 〈 math 〉;) and therein be inscribed C, which is the Result of the absolute term divided by the third part of the Co-efficient, (that is, 〈 math 〉;) And either of the Arches to which this Chord subtends be divided into three equal parts: The Chord which subtendeth to one of those parts is an (Affirmative) Root of that Equation; which therefore hath two Affirmative Roots; suppose A, and E.

LXII. But it hath moreover a Negative Root; which is the Subtense of either of those Arches (whose Chord is A, or E,) increased by a Trient of the whole Circumference, suppose Z. I say either of those Arches; for the same Chord Z, which on the one side subtends a Trient increased by the Arch A, subtends on the other side a like Trient increased by E.

LXIII. But if the Equation be of this Form, Zc − 3RqZ = RqC: The process is just the same in all Points; save, that then, there is but one Affirma∣tive Root, Z; and two Negatives, A, E.

LXIV. But, in both Cases, it must be still observed, That the Chord C be not greater than 2R. (For when this happens, the Chord C, as being greater than the Diameter, cannot be inscribed in such Circle.) Or, (which is in effect the same;) That the Square of Half the Absolute term, be not greater than the Cube of a third part of the Co-efficient of the middle term. For, the third part of that Co-efficient being Rq, and the Cube thereof R6; and half the Abso∣lute quantity ½ RqC, and the Square of this ¼ RqqCq; if this Square be greater than that Cube, and therefore (dividing both by Rqq) ¼ Cq greater than Rq, and (taking the Roots of both) ½ C greater than R; then must C be greater than 2R the Diameter, and therefore cannot be inscribed in the Circle. And therefore, when this happens, such Equations cannot be thus resolved by the Trisection of an Arch. But they may by (what are wont to be called) Cardan's Rules, (as I have elsewhere shewed,) the consideration of which doth not belong to this place.

LXV. If an Arch to be Tripled be a Trient (or Two, Three, Four, or more Trients;) the Triple Arch will therefore be one intire Revolution, (or Two, Three, Four, or more intire Revolutions;) and the Subtense of the Triple will be nothing; (the beginning and end of such Triple Arch being the same Point:) That is, 〈 math 〉, (or 〈 math 〉. And therefore 〈 math 〉, 〈 math 〉, 〈 math 〉. And 3Rq = Aq = Eq = Zq. That is, (as was before shewed)

LXVI. The Square of the Subtense of a Trient, (or of the side of an Equilater inscribed Triangle) is equal to three Squares of the Radius.

Page  13LXVII. If an Arch to be Tripled be a Quadrant; it is manifest that the Sub∣tense* of the Triple, is equal to that of the single. (For the same Chord, sub∣tendeth, on the one side, to Three such Arches; and, on the other side, but to one.) That is, 〈 math 〉; and therefore 〈 math 〉; and 2Rq = Aq. That is,

LXVIII. The Square of the Subtense of a Quadrant (or of the side of an inscribed Quadrate) is equal to two Squares of the Radius.

LXIX. The same may be inferred, from the Bisection of a Semicircumference. For the Subtense of that being 2R; and therefore (by § 9, Chap. preced.) 〈 math 〉 Or, (putting E for the Remainder of that Quadrant to the Semicircle) 〈 math 〉. Or, (because, in this case E = A,) 〈 math 〉. Therefore 2Rq = Aq, as before; and 〈 math 〉.

LXX. But if the Arch to be Tripled be of a Semicircle (and so, greater than a Trient,) the Subtense of the Triple will be the same with that of the single; but with a contrary sign, (by § 28;) and therefore (by § 36.) 〈 math 〉; that is, 〈 math 〉, or Zc = 4ZRq, and Zq = 4Rq, and Z = 2R. Which is the third, or Negative Root, of the last mentioned Equation, 〈 math 〉; beside the two Affirmatives A, E, the Subtenses of the two Quadrants: This being equal to the Aggregate of both, with the contrary sign.

LXXI. Moreover; because the same Subtense (before mentioned,) C = A, subtends not onely to the Triple of a Quadrant on the one side; but also, on the other side, to the single Quadrant; to a third part of this therefore the Root E is to be also a subtendent: (that is, to an Arch of 30 degrees.) That is, 〈 math 〉.

LXXII. And, because (by § 15.) Aq + AE + Eq = 3Rq, and (by § 67.) Aq = 2Rq; therefore AE + Eq = Rq. And therefore (by resolving this E∣quation) 〈 math 〉. And therefore, 〈 math 〉. That is,

LXXIII. As the Subtense of a Quadrant, to the Subtense of a Trient wanting the Radius; so is the Radius, to the Subtense of the Semi-sextant; or, of 30 degrees: Or,

LXXIV. As the side of the (inscribed Equilater) Tetragone, to the Difference of the sides of the Trigone and Hexagone; so is the Radius, to that of the Dode∣cagone. (Understand it of the inscribed Equilater Figures; and so afterward in like cases.)

LXXV. And because (as before) 〈 math 〉: Therefore, 〈 math 〉, or, 〈 math 〉 into Rq. And so, 〈 math 〉 Eq. That is,

Page  14LXXVI. As the Radius, to the Excess of the Diameter above the Subtense of* 120 degrees, (or side of the inscribed Trigone:) so is the Square of the Radius, to that of the Subtense of 30 degrees; or of the side of the Dodecagone. And therefore, (that of Rq to Eq, being Duplicate to that of R to E.)

LXXVII. The Proportion of the Radius, to the Difference of the Diameter and the side of the inscribed Trigone; is Duplicate to that of the Radius, to the side of the inscribed Dodecagone. And therefore,

LXXVIII. The (Radius or) side of the Hexagone, and of the Dodecagone, and the Difference of the Diameter from that of the Trigone, are in continual Pro∣portion.

LXXIX. And (because 〈 math 〉 into 〈 math 〉 into R,) The Excess of the Diameter above the Subtense of the Trient, Multiplied into the Radius; is equal to the Square of the Subtense of 30 degrees, or the Semisextant.

LXXX. The same are found by bisecting the Sextant; (for a quarter of the Trient, or half the Sextant, is the same;) in this manner,

LXXXI. If E be put for the Subtense of 30 degrees, and A for that of the Residue to a Semicircumference, or of 150 degrees; then because the Subtense of a Sextant, or the double Arch of E, is R; therefore (by § 14, Chap. preced.) 〈 math 〉: And Rqq=4RqAq−Aqq=4RqEq−Eqq: And (by resolving that Equation) 〈 math 〉 into 〈 math 〉 into R,=Aq, and Eq. That is,

LXXXII. The Squares of the Subtenses of 150, and of 30 degrees; are equal, that to the Sum, this to the Difference, (of the Diameter and side of the inscribed Tri∣gone) Multiplied by the Radius.

LXXXIII. Again, for as much as C=A subtends as well the Triple of the Arch A of 90 degrees, as the Triple of the Arch E of 30 degrees; therefore 〈 math 〉 is the Subtense of a Trient increased by the Arch A or E; that is, as well of degr. 210 = 120 + 90, as of 150 = 120 + 30. Which was before concluded at § 81. (for Z here, is the same with A there) and Zq (as there Aq,) 〈 math 〉 into Rq.

LXXXIV. And the same is yet again found by subducting 〈 math 〉 into Rq) the Square of the Subtense 30 degrees, out of (4Rq) the Square of the Di∣ameter: (because 150 + 30 = 180 degrees, complete the Semicircumference:) For if from 4Rq we subduct 〈 math 〉, there remains 〈 math 〉, or, 〈 math 〉 into Rq, the Subtense of 150 degrees; and therefore also of 210 degrees.

Page  15

CHAP. III. Of the Quadruplation and Quadrisection of an ARCH or ANGLE.

I. IF in a Circle be inscribed a Quadrilater, whose opposite sides are A, A,* (the subtenses of a single Arch) and B, D, (the subtenses of the Double and Quadruple) the Diagonals will be C, C, (the subtenses of the Triple) as is evident. (But it is evident also, that, in this Case, the Arch A, is less than a Quadrant of the whole Circumference.)

II. And therefore (the Rect-angle of the Diagonals being equal to the two Rect-angles of the opposite sides) Cq−Aq=BD. And therefore 〈 math 〉, and 〈 math 〉 That is,

III. The Square of the Subtense of the Triple Arch, wanting the Square of the Subtense of the single Arch (less than a Quadrant) is equal to the Rect-angle of the subtenses of the Double and Quadruple. And, divided by either of these, it gives the other of them.

IV. But C+A into C−A is equal to Cq−Aq. And therefore, 〈 math 〉 That is,

V. As the Subtense of the double Arch, is to the sum of the Subtenses of the Triple, and of the single (this being less than a Quadrant;) so is the excess of the Subtense of the Triple above that of the single, to that of the Quadruple.

VI. And because (by § 8, Chap. preced.) 〈 math 〉; and therefore 〈 math 〉: Therefore 〈 math 〉.

VII. But (by § 7 and 9, Chap. 29.) 〈 math 〉. And therefore 〈 math 〉. (For if 〈 math 〉 be divided by 〈 math 〉, it is 〈 math 〉: And this again divided by 4Rq−Aq, is 〈 math 〉: But this last Division being by 4Rq−Aq, whereas (according to the value of B) it should be divided only by the Square Root hereof, therefore we are to restore a Multiplication by that Root; which makes it 〈 math 〉.)

VIII. And then, turning the Equation into an Analogy, 〈 math 〉. That is,

Page  16IX. As the Cube of the Radius, to the Subtense of the single Arch (less than a*Quadrant) Multiplied into the double Square of the Radius wanting the Square of the Subtense, so is the Subtense of what this Arch wants of a Semicircumference, to the Subtense of the Quadruple Arch.

X. Or thus; (dividing the two first terms by Rq,) 〈 math 〉. That is,

XI. As the Radius, to the double of the Subtense of the single Arch (less than a Qua∣drant) wanting the Cube of that Subtense divided by the Square of the Radius: So is the Subtense of what that single Arch wants of the Semicircumference, to the Subtense of the Quadruple Arch.

XII. But (by § 8, Chap. preced.) 〈 math 〉. And therefore,

XIII. As the Radius to the excess of the Triple Arch above that of the single (less than a Quadrant:) so is the Subtense of what that single Arch wants of a Semicircum∣ference, to the Subtense of the Quadruple Arch.

XIV. The same may be thus also demonstrated: Because (by § 9, Chap. 29.) 〈 math 〉, is the Subtense of the double Arch of A: Therefore (by the same reason) 〈 math 〉, the Subtense of the double Arch of B; that is, of the Quadruple Arch of A.

XV. And, because 〈 math 〉: Therefore, for 4Rq−Bq, we may put 〈 math 〉, or 〈 math 〉: And the Quadratick Root hereof, 〈 math 〉 Multiplied into 〈 math 〉 makes 〈 math 〉 As before,

XVI. We may also, to the same purpose, (and with the same event,) inscribe* a Quadrilater, so as that A, D, and A, B, may be opposite sides, and B, C, Diagonals. For then BC−BA=DA. And therefore 〈 math 〉 into 〈 math 〉 will equal DA. That is, 〈 math 〉. And 〈 math 〉; as before. But of this we shall say more at § 86. &c.

XVII. But, for as much as the same D, Subtends not only the Quadruple of* the Arch A, but also the Quadruple of the Arch E, (which therefore, together with the Arch A, will complete a Quadrant of the whole Circumference:) it may in like manner be shewed, that 〈 math 〉: And therefore,

XVIII. An Arch less than a Quadrant, and the Arch which this wants of a Qua∣drant, have both the same Subtense of the Quadruple Arch, D. And, accordingly, AE, are two Affirmative Roots of that Equation.

Page  17XIX. But there are yet two other Roots (but both Negative, as will after* appear) of the same Equation; (which we will call P, S;) whereof one sub∣tends a Quadrant increased by the Arch A; the other, a Quadrant increased by the Arch E. For it is manifest (by what is said at § 23, Chap. preced.) that these also must have the same ••btense of the Quadruple Arch, with A and E. For Four times ¼+A, is 1+4, and will therefore have the same Subtense with 4A. And the like of Four times ¼+E, which is 1+4E, whose Sub∣tense is the same with that of 4E. (And the like will follow, in case two, three, or more Quadrants be thus increased.) And consequently,

XX. An Arch greater than a Quadrant, (or than two, three, or more Quadrants) will require the same Subtense of its Quadruple Arch, with its excess above a Quadrant, (or above these two, three, or more Quadrants.)

XXI. But the same P, subtends as well to a Quadrant increased by the Arch of A, as to three Quadrants wanting the said Arch; as also to a Semicircumference (or two Quadrants) increased by the Arch of E, or wanting that Arch. (As is manifest to view by the Scheme.) And, in like manner, S subtends as well to a Quadrant increased by the Arch of E, as to three Quadrants wanting that Arch; as also to a Semicircumference (or two Quadrants) increased by the Arch of A, or wanting that Arch.

XXII. Now, that P, S, are Negative Roots, will thus appear. For, sup∣posing (for instance) 〈 math 〉; and P, a subtendent of an Arch greater than a Quadrant: (But, less than three Quadrants; otherwise it is the same as if it were less than one Quadrant: For the same Chord which subtends an Arch greater than three Quadrants, subtends also to less than one:) P will in this case be greater than 〈 math 〉 the Subtense of a Quadrant, and therefore 2Rq−Pq a Negative quantity (because of greater quantity subtracted from a lesser;) and therefore also P must be Negative, that so 2RqP−Pc (compounded by the Multiplication of two Negatives) may be a Positive quantity, and therefore the whole 〈 math 〉 Affirmative also. (And what is said of P, holds in like manner of S.)

XXIII. But if we chuse to make P Affirmative, then must 〈 math 〉, be Negative: And therefore (changing the signs) 〈 math 〉, Affirmative. (And the like of S.) But, of this, more afterwards.

XXIV. But for what reason, the Equation 〈 math 〉, or 〈 math 〉, hath two Affirmative Roots A, E; and two Negatives P, S; for the same reason will the Equation 〈 math 〉, or 〈 math 〉, have two Negatives A, E; and P, S, Affirmatives.

XXV. If now we consider the Quadrilater, whose said opposite sides are A, A, and E, P; then (because the Arches A, E, do together make up a Quadrant) the Diagonals Q, Q, are subtenses of a Quadrant, (or sides of an inscribed Square) and therefore (by § 68, Chap. preced.) Qq=2Rq, and 〈 math 〉.

Page  18XXVI. And therefore Qq − Aq = 2Rq − Aq = EP; and consequently*〈 math 〉, and 〈 math 〉.

XXVII. But the same P doth also subtend a Semicircumference wanting the Arch of E: And therefore 〈 math 〉 And 〈 math 〉.

XXVIII. And (by the same reason) taking a Quadrilater whose opposite sides are E, E, and A, S; we have the Diagonals 〈 math 〉 And Qq − Eq = 2Rq − Eq = AS. And consequently (because the Arches A and S do complete the Semicircumference) 〈 math 〉; and 〈 math 〉.

XXIX. Now because (as at § 27.) 〈 math 〉; there∣fore 〈 math 〉. And therefore 4RqEq − Eqq = PqEq = 4Rqq − 4RqAq + Aqq; and Aqq + Eqq = 4AqRq + 4EqRq − 4Rqq. (And, in like manner, because 〈 math 〉; therefore 〈 math 〉; and 4RqAq − Aqq = SqAq = Rqq = 4Rqq − 4RqEq + Eqq; and 4AqRq + 4EqRq − 4Rqq = Aqq + Eqq.)

XXX. Now the Legs A, E, contain a Sesquiquadrantal Angle, or of 135 Degrees: (As being an Angle in the Peripherie, standing on an Arch of three Quadrants:) And therefore,

XXXI. In a Right-lined Triangle, whose Angle at the Top is 135 Degrees, if the double of the Aggregate of the Squares of the Legs containing it, (2Aq + 2Eq,) wanting the Square of the Base (Qq = 2Rq,) be Multiplied into the Square of the Base (2Rq;) the Product (4AqRq + 4EqRq − 4Rqq = 2Aq + 2Eq − 2Rq into 2Rq,) is equal to the Biquadrates of the Legs (Aqq + Eqq.) So that,

XXXII. From hence appears, A convenient Method for Adding of Biqua∣drates.

XXXIII. The Subduction of Biquadrates, may (with a little alteration) be performed almost in the same manner. But it is more conveniently done by Multiplying the Sum of the Squares, by the Difference of them. (For Aq + Eq into Aq − Eq is equal to Aqq − Eqq.) But that is a speculation not of this place.

XXXIV. Again, In such Triangle (whose Angle at the Top is of 135 Degrees) If the double of the Aggregate of the Squares of the Legs, be Multiplied into the Square of the Base; the Product is equal to the Aggregate of the Biquadrates of all the sides. For, since 4AqRq + 4EqRq − (Qqq =) 4Rqq = Aqq + Eqq; there∣fore Aqq + Eqq + Qqq = 4AqRq + 4EqRq = 2Aq + 2Eq into (2Rq =) Qq.

Page  19XXXV. Again, because (as at § 27, 28.) 〈 math 〉* and therefore 〈 math 〉, and 4Rqq − 4RqAq + Aqq = PqEq = 4PqRq − Pqq. Therefore Pqq + Aqq = 4PqRq + 4AqRq − 4Rqq. (And in like manner, because 〈 math 〉; and therefore 〈 math 〉, and 4Rqq − 4EqRq + Eqq = SqAq = 4SqRq − Sqq: Therefore, Sqq + Eqq = 4SqRq + 4EqRq − 4Rqq.)

XXXVI. But both A, P, and also E, S, contain a Semiquadrantal Angle, or of 45 Degrees: (As being an Angle in the Periphery standing on a Quadrantal Arch;) And one of the Angles at the Base, Obtuse. And therefore,

XXXVII. In a Right-lined Triangle, whose Angle at the Top is of 45 Degrees, or half a Right-angle (one of the other being Obtuse) If the double of the Aggregate of the Squares of the Legs (as 2Pq + 2Aq) wanting the Square of the Base (Qq = 2Rq) be Multiplied into the Square of the Base (2Rq) the Product (4PqRq + 4AqRq − 4Rqq = 2Pq + 2Aq − 2Rq into 2Rq,) is equal to the Biquadrates of the Legs, (Pqq + Aqq,) In like manner, 2Sq + 2Eq − 2Rq into 2Rq, = 4SqRq + 4EqRq − 4Rqq = Sqq + Eqq. So that

XXXVIII. Here is another Method of Adding Biquadrates.

XXXIX. And likewise; In such Triangle, whose Angle at the Top is of 45 De∣grees, (and one of the other Obtuse,) if the double of the Aggregate of the Squares of the Legs, be Multiplied into the Square of the Base; the Product is equal to the Biqua∣drates of all the sides. For, because 4PqRq + 4AqRq − 4Rqq = Pqq + Aqq; therefore Pqq + Aqq + (4Rqq =) Qqq = 4PqRq + 4AqRq = 2Pq + 2Aq into (2Rq =) Qq. And, because 4SqRq + 4EqRq − 4Rqq = Sqq + Eqq, therefore Sqq + Eqq + (4Rqq =) Qqq = 4SqRq + 4EqRq = 2Sq + 2Eq into (2Rq =) Qq.

XL. Furthermore; If in a Circle be inscribed a Quadrilater, whose opposite* sides are S, A, and Q, Q, and the Diagonals P, P, (as in the Scheme;) Then Pq − (Qq =) 2Rq = SA. And therefore, 〈 math 〉: And 〈 math 〉: And therefore, 〈 math 〉, and 〈 math 〉. And conse∣quently Pqq − 4PqRq + 4Rqq = AqSq = 4RqAq − Aqq = 4RqSq − Sqq.

XLI. And, by the same reason, If a Quadrilater be inscribed whose opposite sides are E, P, and Q, Q; and the Diagonals S, S: Then Sq − (Qq =) 2Rq = EP: And therefore, 〈 math 〉: And 〈 math 〉: And therefore, 〈 math 〉: And 〈 math 〉. And consequently, Sqq − 4SqRq + 4Rqq = PqEq = 4PqRq − Pqq = 4EqRq − Eqq.

XLII. And either way, we may conclude, Sqq + Pqq = 4SqRq + 4PqRq − 4Rqq.

Page  20XLIII. But P, S, contain half a Right-angle, or Angle of 45 Degrees; (as* being an Angle in the Periphery standing on a Quadrantal Arch;) and both the other Angles Acute. And therefore,

XLIV. In a Right-lined Triangle, whose Angle at the Top is 45 Degrees, or half a Right Angle: (and both at the Base, acute:) If the double of the Aggregate of the Squares of the Legs (as 2Sq + 2Pq) wanting the Square of the Base (Qq = 2Rq) be Multiplied into the Square of the Base (2Rq) the Product (4SqRq + 4PqRq − 4qq = 2Sq + 2Pq − 2Rq into 2Rq) is equal to the Biquadrates of the Legs. (Sqq + Pqq.)

XLV. And this is a third Method of Adding Biquadrates.

XLVI. And likewise, in such Triangles, (whose Angle at the Top is of 45 De∣grees, and both the others acute,) if the double of the Aggregate of the Squares of the Legs, be Multiplied into the Square of the Base; the Product is equal to the Biqua∣drates of all the sides. For, since Sqq + Pqq = 4SqRq + 4PqRq − 4Rqq, therefore Sqq + Pqq + (4Rqq =) Qqq = 4SqRq + 4PqRq = 2Sq + 2Pq into 2Rq.

XLVII. These Theorems thus demonstrated severally; whether the Angle at the Top be of 135 Degrees, or of 45 Degrees; and this whether the Triangle be Acute-angled, or Obtuse-angled, (to either of which we may refer the Rect∣angled;) may be thus reduced to these Generals.

XLVIII. In a Right-lined Triangle, whose Angle at the Vertex is either of 135 Degrees, or of 4Degrees; the double Aggregate of the Squares of the Legs contain∣ing it, wanting the Square of the Base, Multiplied into the Square of the Base, is equal to the Biquadrates of the two Legs. (Which is the Addition of Biquadrates.) By § 31, 37, 44.

XLIX. And that double Aggregate of the Squares of the Legs, Multiplied into the Square of the Base, is equal to the Biquadrates of all the three sides. By § 34, 39, 46.

L. Now, the Equation, (at § 29.) 4AqRq + 4EqRq − 4Rqq = Aqq + Eqq (and the other like to it at § 35, 42.) is a Quadratick Equation of a Plain Root: Whereof the Root is 2Rq; the Co-efficient of the middle Term, 2Aq + 2Eq; which is therefore equal to the sum of two Quantities, whose Rect-angle is equal to the Absolute Quantity Aqq + Eqq.

LI. If we therefore order this according to the Rule of other Equations of the same form; and, accordingly, from Aqq + 2AqEq + Eqq (the Square of half the Co-efficient Aq + Eq) we subtract (the Absolute Quantity) Aqq + Eqq; the Remainder is 2AqEq: And the Square Root of this (〈 math 〉) Added to, or Subducted from, half the Co-efficient Aq + Eq, gives the Root of that Equation 〈 math 〉.

LII. But, of this Ambiguous Equation, 'tis evident that we are to make choise of the greater Root, in the case of § 29: Because the Angle at the Vertex (135 Degrees) is greater than a Right-angle; and therefore the Square of the Base (Qq) is to be greater than (Aq + Eq) the Squares of the two sides con∣taining it. And therefore 〈 math 〉. That is,

LIII. If to the Squares of the Legs containing an Angle of 135 Degrees, (or three halfs of a Right-angle,) we add the Rect-angle of those Legs Multiplied by 〈 math 〉; the Aggregate is equal to the Square of the Base.

Page  21LIV. in the same manner may be shewed, that the Equations of § 35. Pqq* + Aqq = 4PqRq + 4AqRq − 4Rqq, (or Sqq + Eqq = 4SqRq + 4EqRq − 4Rqq,) and of § 42. Sqq + Pqq = 4SqRq + 4PqRq − 4Rqq; are Quadratic Equations of a Plain Root 2Rq. But, in all these ('tis manifest) the lesser Root is to be chosen, because the Angle at the Vertex (be∣ing of 45 Degrees) is less than a Right Angle; and therefore the Square of the Base less than the two Squares of the Legs. And therefore, the Root, 〈 math 〉; and 〈 math 〉; and 〈 math 〉. That is,

LV. If from the Squares of the Legs containing an Angle of 45 Degrees (or half a Right-angle,) we subtract the Rect-angle of these Legs Multiplied by 〈 math 〉: the Re∣mainder is equal to the Square of the Base.

LVI. Or we may put both together, thus: If to the Squares of the Legs, bè Added, if they contain an Angle of 135 Degrees; or subtracted thence, if they contain an Angle of 45 Degrees; a Rect-angle of those Legs Multiplied into 〈 math 〉: The Result is equal to the Square of the Base. By § 53, 55.

LVII. We are next to Note, That the subtenses E and P, as also A and S,* (whose two Arches do together make up a Semicircumference,) do (by § 9, Chap. 29.) require the same Subtense of the double Arch: And therefore much more, the same Subtense of the Quadruple. That is, 〈 math 〉 is the Subtense of the double Arch both of E, and of P: And 〈 math 〉, of the double Arch of A, and of S.

LVIII. The Subtense therefore of the Triple Arch of E, (less than a Quadrant, and therefore, much more, less than a Trient,) is 〈 math 〉, (by § 2, 5, Chap. preced.) as being the Square of the Subtense of the double Arch 〈 math 〉, wanting the Square of the Subtense of the single Arch Eq, divided by the Subtense of the single Arch E.

LIX. But the same Subtense of that Triple Arch, (by § 8, 9, Chap. preced.) is 〈 math 〉.

LX. Therefore 〈 math 〉 And 〈 math 〉, the Aggregate of the subtenses of the Triple and single.

LXI. Which may also be thus proved: Because Pq + Eq = 4Rq (as being in a Semicircle,) and therefore Pq = 4Rq − Eq, and PqE = 4RqE − Ec, or PqE − RqE = 3RqE − Ec. Therefore is 〈 math 〉 the Subtense of the Trible; and 〈 math 〉 the Aggregate of the subtenses of the Triple and single.

LXII. And, by just the same reason, 〈 math 〉 is the Subtense of the Triple Arch of A: And 〈 math 〉 the Aggregate of the subtenses of the Triple and single.

Page  22LXIII. Now the Arch of P, (a Quadrant increased by A, its greater Segment)* being greater than a Trient, but less than two Trients; the Subtense of its Triple Arch is 〈 math 〉 (by § 32. Chap. preced.) And 〈 math 〉. (By § 37. Chap. preced.)

LXIV. And therefore, 〈 math 〉 the Subtense of the Triple Arch; and 〈 math 〉, the Difference of the Subtenses of the single and Triple; that is, the Excess of the Subtense of the single above that of the Triple.

LXV. But the Arch S (a Quadrant increased by its lesser Segment E) because it may be either lesser or greater than a Trient, according as the Arch E is less or greater than 30 Degrees; the Subtense of the Triple Arch will be either 〈 math 〉, if the Arch S be less than a Trient; or, if greater, 〈 math 〉. And, accordingly, 〈 math 〉, will be either the Sum or Difference of the subtenses of the single and Triple Arch, according as S is less or greater than a Trient.

LXVI. Moreover; having shewed (at § 2.) that in a Quadruplication of an Arch less than a Quadrant, Cq−Aq=BD, (as wherein the Subtense of the Triple is greater than that of the single; and therefore C, C, Diagonals, and A, A, opposite sides:) Now, if the Arch to be Quadrupled be greater than a Qua∣drant, (but less than three Quadrants) as that of P or S; the Subtense of the single will be greater than that of the Triple. For, supposing the single Arch to be ½ ∓ A (and A less than ¼) the Triple will be ½ ∓ 3A; the Subtense of which will be the same with that of ½ ∓ 3A (for one whole revolution is, in this case, Equivalent to nothing;) and this (so long as A remains less than ¼) will be farther (either in excess or defect) from a Semicircumference (and there∣fore require a less Chord,) than ½ ∓ A.

LXVII. And therefore, in this case, P, P, (or S, S,) become Diagonals, and C, C, opposite sides. And, consequently, Pq−Cq=BD (and Sq−Cq=bD;) And 〈 math 〉. That is,

LXVIII. The Square of the Subtense of an Arch greater than a Quadrant (but less than three Quadrants) wanting the Square of the Subtense of the Triple Arch; is equal to the Rect-angle of the subtenses of the Double and Quadruple. And therefore, divided by one of these, it gives the other.

LXIX. But P+C into P−C, is equal to Pq−Cq. And therefore, 〈 math 〉. (And, in like manner, 〈 math 〉) That is,

LXX. As the Subtense of the double Arch, to the Aggregate of the subtenses of the Triple, and of the single (greater than a Quadrant, but less than three Quadrants;) So is the excess of the Subtense of the single Arch above that of the Triple, to the Subtense of the Quadruple.

Page  23LXXI. Since therefore the Subtense of the Triple Arch P=⅓+A (being* greater than a Trient) is 〈 math 〉; whose Square is 〈 math 〉: If this be taken from Pq, and the Remainder (〈 math 〉) divided by 〈 math 〉 (as at § 7;) the Result is 〈 math 〉. That is, dividing it first by 〈 math 〉, and the Result by 4Rq−Pq, or by −Pq+4Rq, and then restoring a Multiplica∣tion by 〈 math 〉)

LXXII. And therefore (changing the equality in an Anology) 〈 math 〉.

LXXIII. The same will happen if we take the Arch S=⅓+E. For though this may be either greater or less than a Trient, according as E is greater or less than 30 Degrees; and (accordingly) the Triple thereof, either 〈 math 〉, or 〈 math 〉: Yet this doth not alter the case at all; for, either way, the Square of it is the same. And therefore (making the Subduction and Division, as § 71.) 〈 math 〉. And 〈 math 〉. That is,

LXXIV. As the Cube of the Radius, to the Subtense of an Arch greater than a Quadrant (but less than three Quadrants,) Multiplied into the Square of the Subtense, wanting two Squares of the Radius; so is the Subtense of its Difference from a Semicircumference, to the Subtense of the Quadruple Arch.

LXXV. Or thus; 〈 math 〉 Or, 〈 math 〉. That is,

LXXVI. As the Radius, to the Cube of the Subtense of an Arch (greater than a Quadrant, but less than three Quadrants) divided by the Square of the Radius, wanting the double of that Subtense; so is the Subtense of the Difference from a Semicircumference, to the Subtense of the Quadruple Arch.

LXXVII. Or thus, because 〈 math 〉; And also (in case the Arch S be also greater than a Trient) 〈 math 〉. Therefore, 〈 math 〉 (And 〈 math 〉) That is,

LXXVIII. As the Radius, to the Aggregate of the Subtenses of the Triple Arch and of the single (this being greater than a Trient, but less than two Trients,) so is the Subtense of its Difference from a Semicircumference, to the Subtense of the Quadruple Arch.

LXXIX. But if the Arch S (though greater than a Quadrant) be less than a Trient; or greater than two Trients, but less than three Quadrants: That is, 〈 math 〉. And therefore, 〈 math 〉 That is,

LXXX. As the Radius, to the Subtense of an Arch greater than a Quadrant, but less than a Trient (or greater than two Trients, but less than three Quadrants) wanting the Subtense of the Triple Arch; so is the Subtense of its Difference from a Semicircum∣ference, to the Subtense of the Quadruple Arch.

Page  24LXXXI. All which are evident from the Scheme; where the Chord D sub∣tends* the Quadruple of the Arches of A, E, P, and S: And B subtends the double of the Arches E and P, and b the double of the Arches of A and S.

LXXXII. And, in the Quadrilater whose sides B, D, be opposite and Parallel; and C, C, opposite sides; and P, P, Diagonals; Pq−Cq=BD, and 〈 math 〉. And likewise, in the Quadrilater wherein b D are opposite and Parallel; c c opposite sides; and S S Diagonals; Sq−cq=bD, and 〈 math 〉.

LXXXIII. And, in the same Figure, where not only the Arch of P, but of S also, are supposed greater than a Trient; two of the Chords S, S, (as well as P, P,) cut the Chord D.

LXXXIV. But in the other Figure, where the Arch of S is supposed (greater* than a Quadrant, but) less than a Trient; the case is somewhat different. For here b (the Subtense of the double Arch of S) falling on the other side of D (the Subtense of the Quadruple,) the Chord D is not cut by any of the Chords S.

LXXXV. But it comes to the same pass, for these two Chords S S (whether they cut or not cut the Chord D,) being no ingredients of the inscribed Qua∣drilater, (but serve only to shew that b is the Subtense of the double Arch;) it is however, Sq−cq=bD.

LXXXVI. The same things as before, may be yet otherwise demonstrated* (and more commodiously) in this manner; Namely, if instead of the Quadri∣later whose four sides and two Diogonals are A, A, C, C, B, D; we take A, A, B, B, C, D; (taking the subtences of the single and double, twice; but, of the Triple, and Quadruple, once:) with almost the same variety of cases, as before. For,

LXXXVII. If the Subtense of the single Arch be A (or E,) less than a Qua∣drant; then A, B, and A, D, will be opposite sides; and B, C, Diagonals. And therefore, CB−AB=AD. And consequently 〈 math 〉 into 〈 math 〉: equal to AD. That is, 〈 math 〉. And 〈 math 〉, as before. And for the same reason, 〈 math 〉. And 〈 math 〉.

LXXXVIII. If the Subtense of the single Arch be P (or S) greater than a* Quadrant, and even greater than a Trient: (but less than two Trients:) Then B, C, and B, P, (or B, S,) will be opposite sides; and D, P, (or D, S,) Diagonals. And therefore BC+BP=PD, (or BC+BS=SD.) And con∣sequently, 〈 math 〉 into 〈 math 〉 equal to PD. That is, 〈 math 〉. And 〈 math 〉 As before. And, by the same reason, BC+BS=SD (if S also be greater than a Trient) and 〈 math 〉.

Page  25LXXXIX. But if the single Arch be that of S (greater than a Quadrant, but)* less than a Trient; (or P greater than two Trients, but less than three Qua∣drants;) then B, C, and D, S, are opposite sides; and B, S, Diagonals. And therefore, BS−BC=DS. And consequently, 〈 math 〉 into 〈 math 〉. That is, 〈 math 〉. And 〈 math 〉: As before. And in like manner, BP−BC=PD (if the Arch of P be greater than two Trients, which is the same as if less than one;) and 〈 math 〉.

XC. From all which ariseth this General Theorem: The Rect-angle of the Subtenses of the single and of the Quadruple Arch, is equal to the Subtense of the double Multiplied into the Excess of the Subtense of the Triple above that of the single, in case this be less than a Quadrant (or more than three Quadrants;) or, into the Excess of the Subtense of the single above that of the Triple, in case the single be more than a Quadrant but less than a Trient (or more than two Trients, but less than three Quadrants;) or, lastly, into the Sum of the Subtenses of the Triple and single, in case this be more than a Tri∣ent, but less than two Trients. That is, AD:=B into

  • C−A; if the Arch of A be less than a Quadrant, or greater than three Quadrants.
  • A−C; if it be greater than a Quadrant, but less than a Trient; or greater than two Trients, but less than three Quadrants.
  • A+C; if it be greater than a Trient, but less than two Trients.

XCI. And, universally, 〈 math 〉. That is, if the Difference of 2RqA and A c (whereof that is the greater if the single Arch be less than a Quadrant, or greater than three Quadrants; but this if contrary∣wise;) divided by Rc, be Multiplied into〈 math 〉Product is equal to D.

XCII. And therefore, 〈 math 〉. That is,

XCIII. As the Cube of the Radius, to the Solid of the Subtense of the single Arch into the Difference of the Square of it self, and of the double Square of the Radius: So is the Subtense of the Difference of that single Arch from a Semicircumference, to the Subtense of the Quadruple Arch.

XCIV. Now what was before said: (at § 15, Chap. 29.) That the Sub∣tense* of an Arch, with that of its Remainder to a Semicircumference (or of its Excess above a Semicircumference) will require the same Subtense of the double Arch; is the same as to say, that, From any Point of Circumference, two Subtenses drawn to the two ends of any inscribed Diameter, (as A, E,) will require the same Subtense (B) of the double Arch.

XCV. And what is said: (at § 12, 26, Chap. preced.) That the Subtense* of an Arch less than a Trient, and of its Residue to a Trient (as A, E,) and of a Trient increased by either of those, (as Z,) will have the same Subtense of the Triple Arch; is the same in effect with this, that, From any Point of the Circumference, three subtenses drawn to the three Angles of any inscribed (Regular) Trigone (as A, E, Z,) will have the same Subtense (C) of the Triple Arch.

Page  26XCVI. And what is said here: (at § 18, 20.) That the Subtense of an* Arch less than a Quadrant, and of its Residue to a Quadrant, (as A, E,) and of a Quadrant increased by either of these, (as P, S,) will have the same Subtense of the Quadruple Arch: Is the same with this, that, From any Point of the Circumference, Four Subtenses drawn to the four Angles of any inscribed (Regular) Tetragone, (as A, E, P, S,) will have the same Subtense (D) of the Quadruple Arch.

XCVII. But the same holds, respectively, in other Multiplications of Arches; as five Subtenses from the same Point, to the five Angles of an inscribed (Regular) Pentagon; and six, to the six Angles of an Hexagon; &c. Will have the same Subtense of the Arches Quintuple, Sextuple, &c. For they all depend on the same common Principle, That a Semicircumference Doubled, a Trient Tripled, a Quadrant Quadrupled, a Quintant Quintupled, a Sextant Sextupled, &c. Make one entire Revolution; which as to this business, is the same as nothing. And therefore, universally,

XCVIII. From any Point of the Circumference, two, three, four, five, six, or more subtenses, drawn to so many (ends of the Diameter, or) Angles of a (Regular) Polygone of so many Angles, however inscribed, will have the same Subtense of the Arch Multiplied by the number of such ends or Angles. And therefore,

CXIX. An Equation belonging to such Multiplication or Section of an Arch or Angle, must have so many Roots (Affirmative or Negative) as is the Exponent of such Multiplication or Section. As two for the Bisection, three for the Trisection, four for the Quadrisection, five for the Quinquisection: And so forth.

C. And consequently, Such Equations may accordingly be resolved, by such Section of an Angle. As was before noted (at § 61, Chap. preced.) of the Trisection of an Angle.

Page  27

CHAP. IV. Of the Quintuplation and Quinquisection of an ARCH or ANGLE.

I. IF in a Circle be inscribed a Quadrilater, whose sides A, F, (the Subtenses* of the single Arch and the Quintuple,) be Parallel; B, B, (subtenses of the double) opposite: The Diagonals will be C, C, (the subtenses of the Triple,) as is evident from the Figure. But it is evident also, that, in this case, the single Arch must be less than a Quintant (or fifth part) of the whole Circumference.

II. And therefore (the Rect-angle of the Diagonals being equal to the two Rect-angles of the opposite sides,) Cq−Bq=AF. (And by the same reasons cq−bq=EF.) That is,

III. The Square of the Subtense of the Triple Arch, wanting the Square of the Subtense of the double Arch, is equal to the Rect-angle of the Subtenses of the single and of the Quintuple; the single Arch being less than a fifth part of the whole Cir∣cumference.

IV. And therefore, if it be divided by one of them; it gives the other. That is, 〈 math 〉; and 〈 math 〉. (And, in like manner 〈 math 〉; and 〈 math 〉.

V. But C+B into C−B is equal to Cq−Bq. And therefore, 〈 math 〉 That is,

VI. As the Subtense of the single Arch (less than a fifth part of the whole Circum∣ference) to the Aggregate of the subtenses of the Triple and double; so is the Excess of the Subtense of the Triple above that of the double, to that of the Quintuple.

VII. And because (by § 8. Chap. 30.) 〈 math 〉; and therefore 〈 math 〉. And (by § 7. Chap. 29.) 〈 math 〉 Therefore, 〈 math 〉 And 〈 math 〉. That is,

VIII. If, to the Quintuple of the Subtense of an Arch less than a Quintant, wanting the Quintuple of the Cube of the same Subtense divided by the Square of the Radius, be added the Quadricube (or fifth Power) of the same Subtense divided by the Biquadrate of the Radius; the Result is the Subtense of the Quintuple Arch.

IX. The same may be otherwise thus evinced; taking a Quadrilater whose* opposite sides are A, A, and F, C; and the Diagonals D, D. And therefore, Dq−Aq=CF. (And, in like manner, dq−Eq=cF.) That is,

Page  28X. The Square of the Subtense of the Quadruple Arch, wanting the Square of the*Subtense of the single Arch (less than a Quintant,) is equal to the Rect-angle of the subtenses of the Triple and Quintuple. And being divided by either of these, it gives the other of them.

XI. And (because D+A into D−A is equal to Dq−Aq,) 〈 math 〉 (And 〈 math 〉) That is,

XII. As the Subtense of the Triple Arch, to the sum of the subtenses of the Quadruple and of the single (this being less than a Quintant,) so is the Difference of these, to the Subtense of the Quintuple.

XIII. But, (by § 7, Chap. preced.)〈 math 〉. And therefore, 〈 math 〉 Which abated by Aq, leaves 〈 math 〉. And this divided by 〈 math 〉; gives 〈 math 〉. As before,

XIV. The same, is a third way, thus evinced; Inscribing a Quadrilater,* whose opposite sides are A, C, and A, F; and the Diagonals B, D. And there∣fore AC+AF=BD; and BD−AC=AF. (And in like manner, bd−cE=EF.) That is,

15. The Rect-angle of the subtenses of the double and Quadruple Arch, wanting that of the subtenses of the single (being less than a Quintant) and of the Triple; is equal to the Rect-angle of the subtenses of the single and Quintuple. And, being divided by either of these, gives the other of them.

XVI. And therefore, 〈 math 〉 (And 〈 math 〉) That is,

XVII. As the Subtense of a single Arch (less than a Quintant,) to that of the double; so is that of the Quadruple, to the Aggregate of the subtenses of the Triple and Quintuple.

XVIII. But 〈 math 〉. And 〈 math 〉 Therefore, 〈 math 〉 into 〈 math 〉. Likewise, 〈 math 〉; and therefore, 〈 math 〉. And therefore, 〈 math 〉. And 〈 math 〉. As before,

XIX. Or, we may thus compute it: Because 〈 math 〉 (as before;) therefore 〈 math 〉. And therefore, (subtracting 〈 math 〉,) 〈 math 〉. As before,

Page  29XX. The same way, a fourth way, be thus evinced; Inscribing a Quadri∣later* whose opposite sides are B, F, and B, A; and the Diagonals C, D. And therefore BA+BF=CD, and CD−BA=BF. (And, in like manner, cd−bE=BF.) And 〈 math 〉. That is,

XXI. The Rect-angle of the subtenses of the Treble and Quadruple Arches, wanting that of the subtenses of the double and single (this being less than a Quintant,) is equal to that of the subtenses of the double and Quintuple. And being divided by the one, it gives the other.

XXII. And therefore, 〈 math 〉 (And 〈 math 〉) That is,

XXIII. As the subtense of the double Arch, to that of the Triple, so is that of the Quadruple, to the Aggregate of the subtenses of the single (being less than a Quin∣tant) and of the Quintuple.

XXIV. But 〈 math 〉; and 〈 math 〉. Therefore, 〈 math 〉. Likewise, 〈 math 〉: And therefore, 〈 math 〉. And conse∣quently, 〈 math 〉. And (dividing by 〈 math 〉:) 〈 math 〉. As before,

XXV. Or, we may thus compute it: Because 〈 math 〉: Therefore, (dividing by 〈 math 〉;) 〈 math 〉. And 〈 math 〉. As before,

XXVI. Or thus; because BA+BF=CD; and therefore, 〈 math 〉:* And also, 〈 math 〉: And 〈 math 〉: There∣fore, 〈 math 〉; and this Multiplied by 〈 math 〉, makes 〈 math 〉: And 〈 math 〉. As before.

XXVII. But if the Arch to be Quintupled be just the fifth part of the whole Circumference, (and consequently the Quintuple Arch one intire Revo∣lution;) the Subtense of that Quintuple will vanish, or become equal to no∣thing.

Page  30XXVIII. And therefore, in this case, 〈 math 〉.* And so 5RqqA − 5RqAc + Acq = 0; and 5Rqq − 5RqAq + Aqq = 0; or, 5Rqq = 5RqAq − Aqq; or, 〈 math 〉. Which is a Quadratick Equation, whose Root is 〈 math 〉, and the Co-efficient of the middle Term 5R, and the absolute quantity 5Rq.

XXIX. Therefore, (by resolving the Equation): 〈 math 〉.

XXX. Of which ambiguous Equation, the lesser Root is to be chosen, That is, 〈 math 〉; and therefore, 〈 math 〉, the Subtense of a Quintant. That is,

XXXI. The Radius Multiplied into〈 math 〉, is equal to the Subtense of a Quin∣tant, or of 72 Degrees.

XXXII. The same may be otherwise thus inferred: If, in a Circle, be in∣scribed* a Regular Pentagon; whose side A shall be reputed as the Subtense of a single Arch: It's evident that the Subtense of the Duple, and of the Triple, will be the same. (For the same Chord which on the one side, subtends the Duple, doth on the other side, subtend the Triple.) And therefore, 〈 math 〉. And 〈 math 〉. And 〈 math 〉. And there∣fore, 〈 math 〉. And therefore, (as before) 〈 math 〉; and so onward as above.

XXXIII. Now because (as is already shew'd) 〈 math 〉: This therefore will be the Subtense of a Sesquiquintant (or one Quintant and an half, or three tenth parts,) that is, of 108 Degrees: As being that Arch which with the Quintant doth complete the Semicircumference. That is,

XXXIV. The Difference of the Squares of the Subtenses of the Trient, and of the Quintant, divided by the Radius; is equal to the Subtense of the Sesquiquintant, or 108 Degrees. (For 3Rq is the Square of the Subtense of the Trient; and Aq, of the Quintant; and the Difference of these 3Rq − Aq divided by the Radius, is the Subtense.) Or thus,

XXXV. If from the Triple of the Radius 3R, be subducted the Square of the Subtense of a Quintant divided by the Radius; the Remainder is the Subtense of a Sesquiquintant, or 108 Degrees.〈 math 〉.

XXXVI. But the Square of the Subtense of a Quintant so divided, is (as* before) 〈 math 〉; which therefore subtracted from 3R, leaves 〈 math 〉 R the Subtense of 108 Degrees.

Page  31XXXVII. Now, if the Radius be cut in extream and mean proportion, the* greater Segment thereof is 〈 math 〉 (by 11. El. 2.) to which if 1 R be added, we have 〈 math 〉 (the Subtense of 108 Degrees as before;) And therefore,

XXXVIII. If the Radius being cut in extreme and mean proportion, the greater Segment thereof, be added to the whole Radius; the sum is equal to the Subtense of 108 Degrees.

XXXIX. Yet again; If, of a Pentagone so inscribed, the side A be considered as the Subtense of a single Arch; the same will also be the Subtense of the Qua∣druple. (For the same Chord subtends on the one side to one Quintant, and on the other side to four such.)

XL. And therefore, in this case, 〈 math 〉. And RcA = 2RqA − Ac into 〈 math 〉. That is, Rc = 2Rq − Aq, into 〈 math 〉. And (the Square hereof) Rcc = 16Rcc − 20RqqAq + 8RqAqq − Acc; or 15Rcc − 20RqqAq + 8RqAqq − Acc = 0.

XLI. Now this last Equation, if divided by 3Rq − Aq = 0, will afford this Equation; 5Rqq − 5RqAq + Aqq = 0.

〈 math 〉

XLII. And therefore 3Rq = Aq; is one of the Plain Roots of that Equation. And therefore, 〈 math 〉, which is the Subtense of a Trient. (Which is true, because also the Quadruple of a Trient, hath the same Subtense with the single Trient.)

XLIII. But there are also two other Plain Roots included in the Resulting Equation 5Rqq − 5RqAq + Aqq = 0; or 5Rqq = 5RqAq − Aqq. For,

XLIV. The lesser of them is 〈 math 〉; the Square of the Subtense of a Quintant. As before,

XLV. The greater of them is 〈 math 〉; the Square of the Subtense of a double Quintant, or of a Triple, (as we shall see afterward) that is, of 144, or of 216 Degrees. For the Qua∣druple of these also, will have the same Subtense with that of the single. For ⅖ × 4 = ⅗ = 1 + ⅗. And ½ × 4 = 12/5 = 2 + ⅖. Where the Excess above the entire Revolutions (which are here Equivalent to nothing) is ⅗, or 4/5, both which have the same Subtense (as at § 32.) over that of the single Arch; that is, 4 ⅖, or ⅗.

Page  32XLVI. Since therefore (as is shewed) 〈 math 〉, is the Square of* the Subtense of a Quintant; the Square of the Subtense of its Residue to the Semi∣circumference must be 〈 math 〉. Which is therefore the Square of the Subtense of 108 (=180 − 72.) And the Quadratick Root thereof 〈 math 〉 as was also before shewed.

XLVII. And for as much as 〈 math 〉 is the Subtense of 108 Degres, that is of 18 Degrees above a Quadrant; let this Subtense be S, and the Subtense of 18 Degrees, (which is the Excess above a Quadrant) E. Therefore, (by § 54, Chap. preced.) 〈 math 〉. And therefore, 〈 math 〉. And (by resolving that Equation) 〈 math 〉. The lesser of which Roots is here to be chosen, because E is the lesser of the two S, E.

XLVIII. But (as is shewed) 〈 math 〉, and therefore, 〈 math 〉 And 〈 math 〉, and therefore, 〈 math 〉, (half the Square of the Subtense of a Quintant,) whose Square Root is 〈 math 〉. And therefore, (the less Root being here of use) 〈 math 〉, the Subtense of 18 Degrees.

XLIX. The same Arch of 18 Degrees, is also the Complement of a Quintant to a Quadrant. And therefore if the Subtense of a Quintant (or 72 Degrees, being less than a Quadrant,) be called 〈 math 〉; and the Subtense of its Complement to a Quadrant (or of 18 Degrees) E: Then (by § 52. Chap. preced.) 〈 math 〉. And therefore, 〈 math 〉. And (resolving the Equation,) 〈 math 〉.

L. But 〈 math 〉, and therefore, 〈 math 〉 (half the Square of the Subtense of the Sesquiquintant, or 108 Degrees;) and the Square Root thereof 〈 math 〉. And 〈 math 〉. And therefore, 〈 math 〉, the Subtense of 18 Degrees, as before. That is,

LI. The Subtense of the Sesquiquintant, or of 108 Degrees, (that is, the greater Segment of the Radius cut in extream and mean proportion, increased by the entire Radius,) Multiplied into〈 math 〉 (for 〈 math 〉,) wanting the Sub∣tense of the Quintant Multiplied also into〈 math 〉 (for 〈 math 〉 into 〈 math 〉) is equal to the double of the Subtense of 18 Degrees. (And half thereof, equal to that Subtense.) Or,

LII. The Difference of the Subtenses of the Sesquiquintant and of the Quintant, (or of 108 Degrees, and of 72 Degrees) divided by〈 math 〉, is equal to the Subtense of 18 Degrees. That is, that Difference is double in Power to this Subtense, (duplum potest,) or, the Square of that, is double to the Square of this.

Page  33LIII. But, The subtenses of the Quintant and Sesquiquintant, (that is, of 72, and* of 108 Degrees, which together complete the Semicircumference) Multiplied the one into the other, (or the Rectangle of them,) divided by the Radius; is equal to the Subtense of the double Arch of either. For, by § 9, Chap. R) AE (B. That is, of 144, or of 216 Degrees. That is, of the double, or Triple Quintant, (these two having the same Subtense.) That is, 〈 math 〉. That is,

LIV. The Radius Multiplied into〈 math 〉, is equal to the Subtense of the Biquin∣tant, and of the Triquintant; That is, to the Subtense of 144, and of 216 De∣grees.

LV. And the Square of this subtracted from the Square of the Diameter, leaves 〈 math 〉 the Square of the Subtense of 36 Degrees; (as being what 144 Degrees wants of a Semicircumference, and what 216 exceeds it. For 180 − 144 = 36 = 216 − 180.) And the Square Root thereof is that Subtense, 〈 math 〉. That is,

LVI. The greater Segment of the Radius cut in extream and mean Proportion, is the Subtense of 36 Degrees. That is, of half a Quintant, or the side of the inscri∣bed Decagon.

LVII. But we had before shewn (at § 38.) that this added to the Radius (which is the Subtense of 60 Degrees, or side of the inscribed Hexagon,) is equal to the Subtense of 108 Degrees, or Sesquiquintant: Therefore,

LVIII. The Aggregate of the subtenses of 36 Degrees, and of 60 Degrees, (that is, the sides of the inscribed Decagon and Hexagon,) is equal to that of 108 Degrees; (that is, of the Sesquiquintant, or three Tenths.)

LIX. If therefore to the Subtense of 36 Degrees, 〈 math 〉, be added that of 108 Degrees 〈 math 〉, it makes 〈 math 〉, or 〈 math 〉. That is,

LX. The Subtense of the Semiquintant (or 36 Degrees) and of the Sesquiquintant (or 108 Degrees) added together, are in power Quintuple to the Radius, (that is, the Square of that Aggregate is equal to five Squares of the Radius.) For, 〈 math 〉.

LXI. And their Difference is equal to the Radius. For, 〈 math 〉.

LXII. And the Rectangle of them, is equal to the Square of the Radius. For, 〈 math 〉.

LXIII. And the sum of their Squares is Triple to the Square of the Radius. (Or, equal to the Square of the side of the inscribed Trigone.) That is, 〈 math 〉.

Page  34LXIV. And the Difference of their Squares, is in Power Quintuple to the Square of*the Radius, (or, equal to five squared Squares of the Radius. For, 〈 math 〉.

LXV. Again, The sum of the Squares of the subtenses of the Quintant and Biquin∣tant (or of 72 Degrees, and of 144 Degrees,) is Quintuple to the Square of the Radius. For, 〈 math 〉.

LXVI. And the Difference thereof, is in Power Quintuple to the Biquadrate of the Radius. For, 〈 math 〉.

LXVII. And the Rectangle of them, is Quintuple of the Biquadrate of the Radius. For, 〈 math 〉.

LXVIII. We have therefore (as hath been severally demonstrated) these subtenses, answering to their several Arches, or portions of the whole Circum∣ferences, viz.

〈 math 〉

LXIX. By the like method we may find the subtenses of 〈 math 〉, 〈 math 〉, 〈 math 〉, 〈 math 〉, of the whole Circumference: (as likewise of 〈 math 〉, 〈 math 〉, 〈 math 〉, 〈 math 〉,) For the Residue of ⅕ to a Quadrant (or Excess of 〈 math 〉 above a Quadrant) is 〈 math 〉; and therefore the Subtense thereof is 〈 math 〉, (as is shewed before at § 48.) or (which is Equivalent) 〈 math 〉. Or, 〈 math 〉. And the Residue to this to the Semicircumference is 〈 math 〉; whose Subtense there∣fore is 〈 math 〉. Or, 〈 math 〉. Again, the Residue of 〈 math 〉 to a Quadrant (or the Excess of ⅖ above a Quadrant,) is 〈 math 〉; whose Subtense therefore is 〈 math 〉. Or, 〈 math 〉. And the Residue of this to a Semicircumference is 〈 math 〉; whose Subtense therefore is 〈 math 〉. Or, 〈 math 〉. For, in such cases, the same value may be expressed in very different ways. (All which may be easiy proved by computation, in like manner as those before going; and like Corollaries easily deduced from them.) And the Remainders of these to the whole Circum∣ferences (〈 math 〉, 〈 math 〉, 〈 math 〉, 〈 math 〉,) have the same Subtenses with them.

LXX. We have therefore, now, these Subtenses, for the Arches and por∣tions following.

Page  35Degrees of Arches. Portions of the whole. Subtenses.*

〈 math 〉

LXXI. Now if all these Arches be compared with the Trient; and the Sums and Differences of them so compared be observed: We shall thence have a great many more Subtenses, by what is before delivered, at § 15, 47, Chap. 30. As for Example,

LXXII. Suppose the Subtense of 72 Degrees to be 〈 math 〉, and the Subtense of 48 (=120−72) to be E. Then, (by § 15, Chap. 30.) Aq+AE+Eq=3Rq; and therefore, AE+Eq=3Rq−Aq. And (by resolving the Equation) 〈 math 〉.

LXXIII. But 〈 math 〉, and 〈 math 〉. Therefore, 〈 math 〉 the Sub∣tense of 48 Degrees, and therefore also of 312 Degrees.

LXXIV. In like manner: Suppose (as before) A the Subtense of 72 Degrees; and Z the Subtense of 192 (=120+72.) Then (by § 47, Chap. 30.) Zq−AZ+Aq=3Rq; and Zq−AZ=3Rq−Aq. And (resolving the Equa∣tion) 〈 math 〉. The Subtense of 192 Degrees, and therefore also of 168 Degrees. That is,

LXXV. If the Subtense of a Trient, be increased by the greater Segment of such Subtense cut in extream and mean proportion; And thereunto be Added, or taken from it, the Subtense of a Quintant: The result is, in case of Addition, the double Subtense of 168 and of 192 Degrees; in case of Subtraction, the double Subtense of 48 and of 312 Degrees. Or thus,

Page  36LXXVI. If to the greater Segment of the Subtense of a Trient (cut in extream and*mean Proportion,) be added the Sum, or Difference of the subtenses of the Trient and of the Quintant: The Result is, the double Subtense, in the first case, of 168 and of 192 Degrees; in the latter case, of 48 and of 312 Degrees. For, 〈 math 〉, is the half of the Subtense of a Trient (〈 math 〉) increased by its greater Segment is so cut. And 〈 math 〉, is half the Subtense of a Quin∣tant.

LXXVII. And the Squares of these subtenses, subtracted from 4Rq, give us the Squares of the subtenses of their Differences from a Semicircumference. That is, of 12 and of 132 Degrees (whereby 168 and 48 come short of a Semi∣circumference; and whereby 192 and 312 exceed it.) For 12=180−168=192−180; and 132=180−48=312−180.

LXXVIII. Again, suppose the Subtense of a Biquintant, or 144 Degrees, (which is also the Subtense of a Triquintant, or 216 Degrees) being greater than a Trient, to be 〈 math 〉 And the Subtense of 96=216−120=240−144, to be A: And the Subtense of 24=144−120=240−216, to be E. Therefore, (by § 47, Chap. 30.) Zq−ZA+Aq=Zq−ZE+Eq=3Rq; and (Zq being greater than 3Rq,) Zq−3Rq=ZA−Aq=ZE−Eq. And resolving the Equation) 〈 math 〉, the Sub∣tense of 96 Degrees if connected by +; or of 24 Degrees, if by −. (That is, 〈 math 〉, in the first case; and 〈 math 〉, in the latter.) That is,

LXXIX. If to the Subtense of a Biquintant be Added, or taken from it, the greater Segment of the Subtense of a Trient (cut in extream and mean Proportion;) it gives, in the first case, the Subtense of 96, and of 264 Degrees; in the latter, that of 24, and of 336 Degrees.

LXXX. And by these again (by subducting the Squares of their subtenses from 4Rq) we have (the Squares of) the subtenses of their Difference from a Semicircumference, whether in Excess or defect. As of 84=180−96=264−180, and of 156=180−24=336−180.

LXXXI. And if in like manner we compare also the rest of those at § 70, with the Subtense of a Trient; we shall thence have the subtenses of these Arches,

    Degrees of Arches.
  • 120−108=12
  • 120−90=30
  • 120−72=48
  • 120−54=66
  • 120−36=84
  • 120−18=102
  • 120±0=120
  • 120+18=138
  • 120+36=156
  • 120+54=174
    Of the Residue to the Semicircumference.
  • 168
  • 150
  • 132
  • 114
  • 96
  • 78
  • 60
  • 42
  • 24
  • 6
    Of the Residues to the whole Circumference.
  • 348.192
  • 330.210
  • 312.228
  • 294.246
  • 276.264
  • 258.282
  • 240.300
  • 222.318
  • 204.336
  • 186.354

Page  37LXXXII. So that (by these here, and those at § 7.) we have subtenses for* every sixth Degree of the whole Circumference: And consequently the Right sines (as being the half of those subtenses) for every third Degree of the Se∣micircumference. And this by the Solution of Quadratick Equations only, without the help of Cubicks or Superior Equations. And between these may in like manner, be interposed as many more as we please, by the continual Bisection of Arches.

LXXXIII. We return now to pursue the former Inquisition which hath been* intermitted. The Equation formerly proposed at § 7, for the Quinquisection of an Arch, 〈 math 〉; beside the two primary Roots A and E, contains yet three other Roots, (by § 98, 99, Chap. preced.) answering to three other Chords drawn (from the same Point with A and E) to three other Angles of the inscribed Pentagon: Which we shall call L, M, N: Whereof L subtends a Quintant increased by the Arch of A, (or three Quintants increased by the Arch of E;) N subtends a Quintant increased by the Arch of E, (or three Quintants with the Arch of A;) M subtends two Quintants increased by either of those Arches A or E. For every of these Arches, if Quintupled, will have the same Subtense (of the Quintuple) F, as well as the Quintuple of the Arches A or E.

LXXXIV. Of these three, (in case A and E be supposed Affirmative Roots,) L and N will be Negative; but M, Affirmative. But contrarywise, in case A and E be supposed Negative: For then L, N, will be Affirmative, and M Negative. For,

LXXXV. When the single Arch is less than a Quintant (or greater than four*Quintants;) or when it is greater than Two, but less than Three; the Subtense of the Triple Arch will be greater than that of the double. (As is easie to apprehend, or may be proved if need be, in like manner as we have formerly done in like cases; as is after shewed at § 93, &c.)

LXXXVI. And therefore, if A (or E,) be the Subtense of the single Arch; then Cq−Bq=AF, (or cq−bq=EF,) will be an Affirmative quantity, (as at § 2.)

LXXXVII. And, in like manner, if M be the Subtense of the single, (greater* than Two Quintants, but less than Three,) Cq−Bq=MF, will be also Affirmative.

LXXXVIII. But when the single Arch is greater than a Quintant, but less than Two, or greater than Three, but less than Four: The Subtense of the double will be greater than that of the Triple.

LXXXIX. And therefore, if the Subtense of the single be N, then Cq−Bq=NF,* will be Negative.

XC. And in like manner, if it be L; then Cq−Bq=LF will be also* Negative.

XCI. That therefore, L, N, may have Affirmative values (as well as F,) we must put the Equations thus, Bq−Cq=NF; and Bq−Cq=LF. And, if so; then the value of the other three Roots will be Negative.

XCII. But if the single Arch be just a Quintant (or two, three or more Quintants)*the Subtense of the double will be equal to that of the Triple. And therefore, (putting V or X, for the Subtense of the single,) Cq∼Bq=VF=o; or Cq∼Bq=XF=o. The Subtense of the Quintuple (in this case) vanishing to nothing.

Page  38XCIII. Now that, for the Arches A, E, M, the Subtense of the Triple is* greater (or at least not less) than that of the Duple; but contrarywise for the Arches L, N; is easie to apprehend upon a little consideration. For if the Arch A, or E, be to Degrees; B is 20; C, 30: If that be 20; B is 40; C, 60: If A be 40; B is 80; C, 120: If A be 60; B is 120; C, 180. (And hitherto is no doubt, because we are not yet past a Semicircumference; and, till then, as the Arches increase, the Chords increase also; though not when we are past 180 Degrees.) If A be 0; B is 140 = 180 − 40; C, 20 = 180 + 30. So that yet the Chord of C, though past a Semicircumference, is greater than that of B, because nearer to a Semicircumference, or 180 Degrees; (for this doth less exceed it, than that wants of it.) And so 'till we come to 72 Degrees, (or 1/ of the whole) for then B is 144 = 180 − 36, and C, 216 = 180 + 36; where the distance is equal, and accordingly the Chord of the Triple equal to that of the double. But when we be past a Quintant, that of the Triple be∣comes less; for if the single Arch N = /5 + E be 73; B is 146 = 180 − 34; C, 219 = 180 + 39; and this doth therefore more exceed 180, than the other comes short of it; and hath therefore the shorter Chord. So likewise, if N be 80; B is 160 = 180 − 20; C, 240 = 180 + 60: If N be 90, B is 180; C, 270 = 180 + 90: If N be 100; B is 200 = 180 + 20 = 360 − 160; C, 300 = 180 + 120 = 360 − 60: Where the Triple is farther from a Semi∣circumference, as more exceeding it; and nearer to a whole Revolution (which is Equivalent to nothing) as approaching nearer to it; and therefore the Chord of the Triple, less than that of the double: So, if N, or L be 108; B is 216 = 180 + 36 = 360 − 144; C, 324 = 360 − 36 = 180 + 144. If L = ⅕ + A be 120; B is 243 = 180 + 60 = 360 − 120; C, 360. And therefore that B, the greater Chord: And so it will be 'till we come to 144 Degrees (or ⅖) when again they will become equal; for then B will be 288 = 180 + 108 = 360 − 72; and C = 432 = 360 + 72 = 540 − 108; which doth as much surpass a whole Revolution as the other wants of it; and doth as much want of a third Semi∣circumference as the other exceeds the first; and therefore their Chords become equal. But after this, the Chord of the Triple doth again become the greater: For if M the single Arch be 145; B will be 290 = 180 + 110 = 360 − 70; C, 435 = 360 + 75 = 540 − 105: If M be 150; B is 300 = 360 − 60; C, 450 = 360 + 90: If M be 180; B is 360; C, is 540 = 360 + 180: If M be 200; B is 400 = 360 + 40; C, 600 = 360 + 240; where the Arch C (as farther remote from an intire Revolution) requires the greater Chord. And so onward 'till we come to 216, (or ⅗) where the Chords of B and C do again become equal, for B will be 432 = 360 + 72; C, 648 = 720 − 72; where the Arch B doth as much exceed one Revolution, as C wants of two; and therefore require equal Chords. After this, the Arches L, N, from 216 to 288, have the same Chords with those of L, N, from 144 backward to 72, (as being their Complements to a whole Revolution,) and the same Chords of their Doubles and Triples, with the Doubles and Triples of those; and there∣fore (as there) the Chords of the Duple greater than those of the Triple. And from thence to 360 (which is an entire Revolution) the Chords are the same with those of A and E, (as being the Remainders of these to an intire Revolu∣tion) and therefore here also, the Chord of the Triple is greater than that of the Duple.

XCIV. All which depends on this General Consideration; (which equally serves for all such Comparings of Arches and their Subtenses; and is therefore to be taken notice of, once for all.) That is,

XCV. Arches equally distant from the beginning or end of (one or more) entire Revolutions, have equal Subtenses, (for the same Chord doth indifferently subtend both or all of them;) But those which are less distant from such beginning or end, have the lesser subtenses; (as nearest approaching to nothing.)

Page  39XCVI. Again, Arches equally distant (whether in Excess or defect) from* 1, 2, 3, (or any odd number of) Semicircumferences, have equal subtenses, (for here also the same Chord subtends both or all;) but those which are less distant from such Semicircumferences, have the greater Subtense, (as nearest approaching to that of a Semicircumference, or 180 Degrees, the greatest Chord of all.)

XCVII. 'Tis manifest therefore, that, if the Arch E or A be not grea∣ter than 60 Degrees, and consequently the Triple Arch do not exceed one Semicircumference, That of the Treble (as nearest approaching to it) will be greater than that of the Double. And though A be greater than 60 Degrees; that of the Triple will yet be the greater, 'till this do as much exceed a Se∣micircumference as the Double comes short of it: That is, 'till 2 ½ A = 180 Deg. or ½ of the whole Circumference; that is, 'till A = ⅕, or 72 Degrees. And what is said of E and A less than /5, doth equally hold of ⅘ + A = 1 − E, and ⅘ + E = 1 − A, which have the same Chords with E and A; and their Double and Treble, the same with the Double and Treble of E and A.

XCVIII. But if N, or L, the single Arch exceed ⅕, suppose ⅕ + E or ⅕ + A; the Subtense of the double will be the longer. For the Subtense of /5 being the same with that of ⅗ = 1 − ⅖; that of 2 N = ⅖ + 2 E will be longer than it, as nearer approaching to ½; ('till 2 N or 2 L = ½, that is, N or L = ¼ or 90 Degrees;) but that of 3 N = /5 + 3 E less than it, as nearer approaching to 1 intire Revolution. And even when 2 L exceeds ½, yet 3 L will have the less Chord, as nearer approaching to 1 intire Revolution; 'till it become equal to it; that is, 3 L = 1, and L = ⅓, or 120 Deg. And even after this, 'till 3 L do as much exceed 1, as 2 L comes short of it; that is, 'till 2 ½ L = 1; or L = ⅖ or 144 Degrees. But then (as before at A or N = ⅕) the Chords will be equal; for then the double is ⅘ = 1 − /5; the Treble 6/5 = 1 + ⅕. And what is said of N = ⅕ + E, or L = /5 + A; holds equally true of N = /5 + A, or L = /5 + E; (that is of 1 − N, or 1 − L;) as having the same Chords with those.

XCIX. But when M the single Arch exceeds /5, suppose ⅖ + E; the Chord of the Treble will again be longer than that of the Double. For the Treble of ⅖ as much exceeding, as the Double of it comes short of, 1 Revolution; the Treble of ⅖ + E will more exceed it, (approaching nearer to the third Semi∣circumference) and the Double want less of it, (approaching nearer to 1 Re∣volution,) 'till 3 M = /2; that is, M = ½ or 180 Degrees. And what is said of M = ⅖ + E less than ½; holds also of M = ⅖ + A = /5 − E: Which doth as much exceed a Semicircumference, as the other comes short of it.

C. 'Tis manifest therefore, that for the Arches A, E, less than ⅕, or more than ⅘ (but less than 1 Revolution;) and again for the Arch M, more than ⅖ but less than /5; the Chord of the Triple is greater than that of the Double; by § 97, 99. But, for the Arches L or M, more than ⅕ but less than ⅖; or more than ⅕, but less than ⅘; the Chord of the Double is greater than that of the Treble; by § 98. But in case the single Arch be ⅕, ⅖, /5, ⅘; (or any number of Quintants,) the Chords of the Double and Treble are equal. And the same method may be pursued in other like Comparisons of Arches and Chords.

CI. Now, (to return where we left off at § 92.) what hath been particularly delivered, may be Collected into this General. Namely, (putting O for the Subtense of the single Arch) C q ∼ B q = O F (by § 85, 88, 92.) And 〈 math 〉. And 〈 math 〉. That is,

Page  40CII. The Difference of the Squares of the subtenses of the Triple and double Arches,*is equal to the Rect-angle of the subtenses of the single and Quintuple. And that Difference applied to either of these, gives the other. (Which is a General to that of § 3.) Namely, if O be interpreted of A, E, M; then C q − B q = O F, and 〈 math 〉. If, of L, N; then B q − C q = O F, and 〈 math 〉. If, of V, X; then B q ∼ C q = V F = 0. And 〈 math 〉. Or, B q ∼ C q = X F = 0, and 〈 math 〉.

CIII. Again, because C q ∼ B q = C + B into C ∼ B; therefore, O. B + C :: B ∼ C. (interpreting C ∼ B, of C − B, for A, E, M; but of B − C, for L, N.) That is,

CIV. As the Subtense of the single Arch, to the Aggregate of the subtenses of the Double and Triple; so is the Difference of these, to that of the Quintuple. (Which is a General to that of § 6.)

CV. But (by § 45, Chap. 30.) 〈 math 〉; and therefore, 〈 math 〉: And (by § 7, Chap. 29.) 〈 math 〉; and therefore 〈 math 〉: From hence therefore, we may have the value of C q ∼ B q = O F, and of 〈 math 〉, sutable to each case. Namely,

CVI. If the Arch O be less than ⅕, or more than ⅘; (that is, from 0, to 72° and from 288, to 360.) Or more than ⅖ but less than ⅗, (that is, from 144, to 216°) there is 〈 math 〉: And 〈 math 〉. And O to be understood of A, E, and M.

CVII. But if the Arch O be more than ⅕ but less than ⅖; or more than ⅗ but less than ⅘; (that is, from 72 Degrees to 144, and from 216 to 288:) Then is, 〈 math 〉: And 〈 math 〉. And O to be interpreted of L, N.

CVIII. That is, (to reduce all to a brief Synopsis)〈 math 〉And, in the common term (or Point of connexion) of these Intervals, it is indifferent to whether of the two to refer them: As at 36 Degrees, to E or A; at 72° to A or N; and so of the rest.

Page  41CIX. Hence follows this Five-fold Equation; containing five Roots. 5 R q q E* − 5 R q E c + E q c = 5 R q q A − 5 R q A c + A q c = (R q q F =) − 5 R q q N + 5 R q N c − N q c = − 5 R q q L + 5 R q L c − L q c = + 5 R q q M − 5 R q M c + M q c.

CX. Now because 5 R q q A − 5 R q A c + A q c = 5 R q q E − 5 R q E c + E q c; therefore, (by transposition) 5 R q q A − 5 R q q E = 5 R q A c − 5 R q E c − A q c + E q c: And (dividing all by A − E,) 〈 math 〉.

CXI. But (by § 13, 14, 15, Chap. 30.) 〈 math 〉, and therefore, 〈 math 〉: Therefore, 〈 math 〉: That is, 〈 math 〉.

CXII. And again, because (as will appear by Division) 〈 math 〉: Therefore, A q q + A c E + A q E q + A E c + E q q = 10 R q q.

CXIII. But the Angle contained by A, E, is of 144 Degrees. (As being an Angle in the Circumference insisting on an Arch of 288 Degrees, or ⅘ of the whole.) Therefore,

CXIV. The Difference of the Quadricubes of the Legs containing an Angle of 144 Degrees, or divided by the Difference of those Legs, is equal to Ten Biquadrates of the Radius of the Circumscribed Circles. (by § 111.) That is, (by § 112.)

CXV. The Biquadrates of the Legs containing an Angle of 144 Degrees, together with the three means proportional between these Biquadrates, is equal to Ten Biquadrates of the Radius of the Circumscribed Circle.

CXVI. But now the Base of this Triangle, being the side of an Inscribed E∣quilater Pentagon, or Subtense of 72 Degrees; is, (by § 31.) 〈 math 〉; and therefore the Square of this 〈 math 〉; and, its Biquadrate, 〈 math 〉: Which is, to 10 R q q, as 〈 math 〉 to 10; or as 〈 math 〉. Therefore,

CXVII. The Difference of the Quadricubes of the Legs containing an Angle of 144 Degrees, divided by the Difference of those Legs, or the Biquadrates of the Legs containing such Angle, together with the three means Proportional between these Bi∣quadrates; is, to the Biquadrate of the Base subtending that Angle; as〈 math 〉.

CXVIII. Again, because (by § 109.) 5 R q q M − 5 R q M c + M q c = (R q q F =) 5 R q q A − 5 R q A c + A q c (M being greater than A;) there∣fore (as at § 110, 111, 112,) 〈 math 〉. And 〈 math 〉:

Page  42CXIX. And (by the same reason) 〈 math 〉.* And 〈 math 〉.

CXX. And also, because (by § 109.) − 5RqqL + 5RqLc − Lqq = (RqqF =) − 5RqqN + 5RqNc − Nqc; and (changing all the signs) 5RqqL − 5RqLc + Lqc = 5RqqN − 5RqNc + Nqc (L being greater than N:) Therefore, (as at § 118.) 〈 math 〉. And 〈 math 〉.

CXXI. But the Angles contained by M, A; and M, E; and L, N; are of 72 Degrees; (as being Angles in the Circumference insisting on an Arch of 144 Degrees, or ⅖ of the whole:) And, as to M, E; one of the Angles at the Base, obtuse; but, as to M, A; all acute: (This being an Angle in a greater Segment; that, in a less, than a Semicircle;) and likewise, as to L, N, all acute. There∣fore, (as at § 114, 115.)

CXXII. The Difference of the Quadricubes of the Legs containing an Angle of 72 Degrees, divided by the Difference of those Legs; is equal to Ten Biquadrates of the Radius of the Circumscribed Circle. And,

CXXIII. The Biquadrates of the Legs containing an Angle of 72 Degrees, together with the three means Proportional between those Biquadrates, is equal to Ten Biquadrates of the Radius of the Circumscribed Circle.

CXXIV. But here the Base of this Triangle (subtended to that Angle of 72 Degrees,) is the Subtense of a Biquintant, or Triquintant; that is, of ⅖ = 144 Degrees; or of ⅗ = 216 Degrees; which is (by § 54.) 〈 math 〉. And the Square of this 〈 math 〉 And its Biquadrate 〈 math 〉. Which is, to 10Rqq, as 〈 math 〉 to 10; or, as 〈 math 〉 to 4. Therefore,

CXXV. The Difference of the Quadricubes of the Legs contanining an Angle of 72 Degrees divided by the Difference of those Legs; or, the Biquadrates of the Legs con∣taining such Angle, together with the three means Proportional between those Biqua∣drates; is, to the Biquadrate of the Base subtending that Angle; as 4 to〈 math 〉.

CXXVI. Again, because (by § 109,) 5RqqA − 5RqAc + Aqc = (RqqF =) − 5RqqL + 5RqLc − Lqc (L being greater than A:) Therefore, (by transposition) 5RqqL + 5RqqA = 5RqLc + 5RqAc − Lqc − Aqc. And (dividing all by L + A,) 〈 math 〉.

CXXVII. But (by § 46, 47, Chap. 30.) 〈 math 〉, and therefore, 〈 math 〉 Therefore, 〈 math 〉. That is, 〈 math 〉.

Page  43CXXVIII. And again, because (as will appear by Division) 〈 math 〉:* Therefore, Lqq − LcA + LqAq − LAc + Aqq = 10Rqq.

CXXIX. But the Angle contained by L, A, is of 36 Degrees (as being an Angle at the Circumfererence insisting on an Arch of 72 Degrees, or ⅗ of the whole,) and one of the other, obtuse.

CXXX. And the same is to be said (for the same reasons) of N, E, as of L, A.

CXXXI. And also, because in like manner (by § 109,) 5RqqM − 5RqMc + Mqc = (RqqF =) − 5RqqN + 5RqNc − Nqc; (M being greater than N:) Therefore, (by the same methods,) 〈 math 〉. And the Angle contained by M, N, is of 36 Degrees; and one of the other, obtuse.

CXXXII. And just the same (for the same reasons) of M, L; save that here the Angles be all acute.

CXXXIII. And these are all the cases that can happen, the Angle at the Vertex being 36 Degrees; for that of the Legs V, X; is to be reduced to that of A, L; and that of X, X; to that of L, M; (and the like is to be understood of other like cases, where A is extended to the whole Quintant, and E vanisheth into nothing.) Therefore,

CXXXIV. The Sum of the Quadricubes of the Legs containing an Angle of 36 Degrees, divided by the Sum of those Legs, is equal to Ten Biquadrates of the Radius of the Circumscribed Circle. (By § 127, 130, 131, 132.) And,

CXXXV. The Biquadrates of the Legs containing an Angle of 36 Degrees, with a mean Proportional between those Biquadrates, wanting the first and third of three means Proportional between them; are equal to Ten Biquadrates of the Radius of the Circumscribed Circle.

CXXXVI. But the Base subtended to this Angle of 36 Degrees, being the side of an inscribed Equilater Pentagon; (as at § 116,) the Biquadrate hereof is to 10Rqq as 〈 math 〉 to 4. And therefore,

CXXXVII. The Sum of the Quadricubes of the Legs containing an Angle of 36 Degrees, divided by the Sum of those Legs: Or, the Biquadrates of the Legs con∣taining such Angle, with a mean Proportional between those Biquadrates, wanting the first and third of three mean Proportionals between them; is, to the Biquadrate of the Base subtending that Angle; as 4, to〈 math 〉.

CXXXVIII. Again, because (by § 109.) 5RqqA − 5RqAc + Aqc = (RqqF =) − 5RqqN + 5RqNc − Nqc; (N being greater than A;) Therefore, (as at § 126, &c.)〈 math 〉.

CXXXIX. And, in like manner, because 5RqqE − 5RqEc + Eqc = (RqqF =) − 5RqqL + 5RqLc − Lqc, (L being greater than E:) There∣fore, 〈 math 〉.

Page  44CXL. But the Angles contained by N, A; or by L E; are Angles of 108* Degrees, (as being Angles at the Circumference, insisting on an Arch of 216 Degrees, or ⅗ of the whole.) Therefore,

CXLI. The Sum of the Quadricubes of the Legs containing an Angle of 108 Degrees, divided by the Sum of those Legs, is equal to Ten Biquadrates of the Radius of a Circumscribed Circle. And,

CXLII. The Biquadrates of the Legs containing an Angle of 108 Degrees, with a mean Proportional between those Biquadrates, wanting the first and third of three means Proportional between them; are equal to Ten Biquadrates of the Radius of a Circum∣scribed Circle.

CXLIII. But the Base subtended to this Angle of 108 Degrees, is the Sub∣tense of a Biquintant, or (which is the same) of a Triquintant; that is, of ⅖ or ⅗ of the whole Circumference: And therefore, (as at § 124) is to 10Rqq, as 〈 math 〉 to 4. Therefore,

CXLIV. The Sum of the Quadricubes of Legs containing an Angle of 108 Degrees, divided by the Sum of those Legs: Or, The Biquadrates of the Legs containing such Angle, with a mean Proportional between those Biquadrates, wanting the first and third of three means Proportional between them; is, to the Biquadrate of the Base subtending that Angle; as 4 to〈 math 〉.

CXLV. Now these several Theorems thus delivered in particular, may be Collected into these Generals following. Namely,

CXLVI. The Difference of the Quadricubes of Legs containing an Angle of 144 or of 72 Degrees, divided by the Differences of those Legs: Or, The Sum of the Quadricubes of Legs containing an Angle of 36 Degrees, or of 108 Degrees, divided by the Sum of those Legs: Or, (which is Equivalent to those) The Biquadrates of the Legs (in the former case) with the three means Proportional between them; Or, (in the latter case) The Biquadrates of the Legs, with a mean Proportional between them, wanting the first and third of three means Proportionals: Are equal to Ten Biquadrates of the Radius of a Circumscribed Circle. And, These, to the Biquadrates of their respective Bases subtending such Angle of 144 Degrees, or of 36 Degrees; are as 4 to 〈 math 〉; but, of the Bases subtending such Angle of 72 Degrees, or of 108 Degrees; as 4 to〈 math 〉: Or, as 8 to 〈 math 〉, and 8 to 〈 math 〉. That is, in the Duplicate proportion 〈 math 〉 to 〈 math 〉; and of 〈 math 〉 to 〈 math 〉.

CXLVII. And those sides, contain these following Angles.

Sides.
Deg.
A, E.
144
  • M, A.
  • M, E.
  • L, N.
72
  • L, A.
  • N, E.
  • M, L.
  • M, N.
36
  • N, A.
  • L, E.
108.

whereof the Four first Couple, are sides of like signs; the six latter, of unlike.

CXLVIII. The same Equations may be thus also considered. Because (by § 110) 5RqqA − 5RqqE = 5RqAc − 5RqEc − Aqc + Eqc: Therefore, (dividing all by A − E, and again by 5Rq,) 〈 math 〉 And (by transposition)

Aq + AE + Eq − Rq, into 5Rq, = Aqq + AcE + AqEq + AEc + Eqq.

Page  45CXLIX. And in like manner (because M, A, and M, E, and L, N, have* also like signs.)

  • Mq+MA+Aq−Rq, into 5Rq, = Mqq+McA+MqAq+MAc+Aqq.
  • Mq+ME+Eq−Rq, into 5Rq, = Mqq+McE+MqEq+MEc+Eqq.
  • Lq+LN+Nq−Rq, into 5Rq, = Lqq+LcN+LqNq+LNc+Nqq.

CL. And therefore, In a Right-lined Triangle, whose Legs contain an Angle of 144 Degrees, (as A, E,) or 72 Degrees, (as M, A, or M, E, or L, N,) if the Squares of the Legs, with the Rectangle of them, wanting the Square of the Radius of the Circumscribed Circle, be all Multiplied into five times the Square of that Radius: The Product is equal to the Biquadrates of the Legs, with three means Proportional between those Biquadrates.

CLI. In like manner may be shewed, (where the signs of the Legs be un∣like,) That,

  • Lq−LA+Aq−Rq, into 5Rq, = Lqq−LcA+LqAq−LAc+Aqq.
  • Nq−NE+Eq−Rq, into 5Rq, = Nqq−NcE+NqEq−NEc+Eqq.
  • Mq−ML+Lq−Rq, into 5Rq, = Mqq−McL+MqLq−MLc+Lqq.
  • Mq−MN+Nq−Rq, into 5Rq, = Mqq−McN+MqNq−MNc+Nqq.
  • Nq−NA+Aq−Rq, into 5Rq, = Nqq−NcA+NqAq−NAc+Aqq.
  • Lq−LE+Eq−Rq, into 5Rq, = Lqq−LcE+LqEq−LEc+Eqq.

CLII. And therefore, In a Right-lined Triangle, whose Legs contain an Angle of 36 Degrees, (as L, A, or N, E, or M, L, or M, N;) or 108 Degrees, (as N, A, or L, E;) if the Squares of the Legs; wanting the Rectangle of them, and the Square of the Radius of the Circumscribed Circle, be all Multiplied into five times the Square of that Radius; the Product is equal to the Biquadrates of the Legs, and a mean Proportional between those Biquadrates, wanting the first and third of three means Proportional between them.

CLIII. Now all this variety of cases, and Deductions from them; from* § 83, hitherto, ariseth from the first Construction, at § 1. and what is Analo∣gous thereunto: Where the six Lines, for the four sides and two Diagonals of the Quadrilater, are F, A; B, B; C, C. And the variety ariseth from hence, that sometimes C, C, are the Diagonals; and B, B, opposite sides; sometimes C, C, are opposite sides; and B, B, Diagonals; according as C or B happens to be greater.

CLIV. But, by a like method, with some little alteration, we may infer most* of the same things; (and observe thence like Deductions, or others Analogous thereunto; with like variety of cases;) from the second Construction, at § 9; where the six Lines are, F, C; A, A; D, D. Where the variety of cases pro∣ceedeth from hence, that sometimes D, D, are Diagonals, and A, A, opposite sides; sometimes D, D, are opposite sides; and A, A, (or what answers to them) Diagonals; according as D, or A, (or what answers to this, E, L, M, N,) are greater.

CLV. And accordingly, the propositions at § 10, and 12, may be delivered more generally. Namely,

Page  46CLVI. The Difference of the Squares of the subtenses of the Quadruple and of the*single Arch; is equal to the Rectangle of the subtenses of the Triple and Quintuple. And, being divided by either of these, it gives the other. And,

CLVII. As the Subtense of the Triple Arch, to the Sum of the subtenses of the Qua∣druple and of the single; so is the Differences of these, to the Subtense of the Quintuple. Whether such single Arch be lesser, or greater, or equal to a Quintant.

CLVIII. And in like manner, from the Third Construction, at § 14. where* the six Lines are, F, A; C, A; B, D. And what is there delivered (at § 15, 17.) of an Arch less than a Quintant, may be more generally delivered, thus,

CLIX. The Difference of the Rectangles of the subtenses of the double and of the Quadruple Arch; and, of the single and Triple; is equal to that of the Subtenses of the single and Quintuple. And, being divided by either of these, it gives the other. And,

CLX. As the Subtense of the single Arch, to that of the double; so is that of the Quadruple, to the Sum or Difference of the subtenses of the Triple and Quintuple; according as B, D, happen to be Diagonals or opposite sides

CLXI. And in like manner, from the Fourth Construction, at § 20; where* the six Lines are, F, B; A, B; C, D. And what is there delivered at § 21, 23, may be more generally delivered; thus,

CLXII. The Difference of the Rectangles, of the subtenses of the Triple and Qua∣druple Arches; and, of the single and double; is equal to that of the subtenses of the double and Quintuple. And, being divided by either, gives the other of them. And,

CLXIII. As the Subtense of the double Arch, is to that of the Triple; so is that of the Quadruple, to the Sum or Difference of the subtenses of the Quintuple, and single; according as C, D, happen to be Diagonals, or opposite sides.

CLXIV. And from every of these Constructions, may be derived like varietie of cases and Consequences, (with Figures suited to those cases:) as (at § 83, &c.) is done from the first Construction. But I forbear to pursue these any farther; and leave it to any who shall think fit, (for their own Exercise,) to pursue these as I have done the first.

Page  47

CHAP. V. Of the Sextuplation, and Sextisection of an ARCH or ANGLE: And other following Multiplications and Sections.

I. ACCORDING to the same methods may be had, the Sextupla∣tion, Septuplation, and other consequent Multiplications; as also the Sextisection, Septisection, and other consequent Sections, of an Arch or Angle. Of which I shall briefly touch at some.

II. The Sextuplation, may be had, by Tripling the Double, or Doubling the Triple Arch. And, accordingly, the Sextisection, by Bisecting the Subtriple, or Trisecting the Subduple. (as is of it self manifest.) And the same holds, in like manner, for Multiplications and Sections which take their Denomination from a Compound number. For Multiplications and Sections successively made, according to the Components of such Compound number, amount to the same as one by such Compound number.

III. But though Six were not a Compound number, or be not considered as such; yet may such Sextuplation and Sextisection be had in like manner as those before. Namely,

IV. If in a Circle be inscribed a Quadrilater, whose opposite sides are B, B,* subtenses of the Duple; and B, G, subtenses of the Duple and Sextuple; and the Diagonals D, D, subtenses of the Quadruple. Then is, Dq−Bq = BG; and B) Dq−Bq (G.

V. Or, Let the opposite sides be A, A, and D, G; and the Diagonals F, F.* Then is, Fq−Aq = DG; and D) Fq−Aq (G.

VI. Or, Let the opposite sides be A, B, and C, G; and the Diagonals D, F.* Then is, DF−AB = CG; and C) DF−AB (G.

VII. Or, Let the opposite sides be A, C, and B, G; the Diagonals C, F.* Then CF−AC = BG; and B) CF−AC (G.

VIII. And therefore, Dq−Bq = CF−CA.

IX. Or, Let the opposite sides be A, G, and A, D; the Diagonals B, F.* Then BF−AD = AG; and A) BF−AD (G.

X. Or, Let the opposite sides be B, C, and A, G; the Diagonals C, D.* Then CD−BC−AG; and A) CD−BC (G.

XI. And therefore, BF−AD = CD−BC.

XII. It is manifest that from hence may be deduced a great number of Equations, and Analogies, and great variety of Theorems, in like manner, as is done in the Chapters foregoing. But I forbear here to pursue them in particular as is there done.

Page  48XIII. But from every of those Constructions, (the values of B, C, D, F,* being known as is above declared,) we have (by ordering the Equations in due manner,) 〈 math 〉. Or, 〈 math 〉. And (taking the Squares of these,) 4GqRcccc−GqRqqccAq = 144RccccAq−456RqqccAqq + 553RqccAcc−328RccAqcc + 102RqqAqqcc−16RqAcccc + Aqcccc.

XIV. That is, (dividing all by 4Rq−Aq,) RqqccGq = 36RqqccAq−105RqccAqq + 112RccAcc−54RqqAqcc + 12RqAqqcc−Acccc.

XV. Of this Equation there be Six plain Roots, answering to Aq; the* Square Roots of which, are A. Which are so many streight Lines from some one Point of the Circumference, to the Six Angles of an inscribed regular Hexagon. (So that, any one of them being known, the rest are known also. And the like in all such Equations.)

XVI. Of these, the Two least, A, E, (which subtend, on the one side, to Arches less than a Sextant; and, on the other side to more than five Sextants;) And the Two greatest, x, y, (which subtend to Arches greater than two Sex∣tants, but less than four;) are Affirmative Roots; (because the Subtendent of the double Arch is less than that of the Quadruple; and therefore Dq−Bq an Affirmative Quantity:) But the Two betwen them I, K, (which subtend on the one side, to Arches greater than one Sextant but less than two; and on the other side, to Arches greater than four Sextants but less than five;) are Nega∣tives, (because of D less than B; and therefore Dq−Bq a Negative Quan∣tity;) G being in all, reputed Affirmative.

XVII. If a Chord be subtendent to just a Sextant, or two or more Sextants; it is indifferent to whether of the two cases on either side it be referred; suppose 〈 math 〉. (which is to be understood in all cases of like nature.) And when ever this happens, one of the Roots vanish, or become equal to nothing.

XVIII. For the Septuplation or Septisection of an Arch or Angle; we shall have, according as the Quadrilater may be differently inscribed, the Subtense of the Septuple Arch, 〈 math 〉, or 〈 math 〉, or 〈 math 〉, or 〈 math 〉, or 〈 math 〉, or 〈 math 〉, or 〈 math 〉, or 〈 math 〉, or 〈 math 〉.

XIX. From every of which Equations, (having the values of B, C, D, F, G, known as before,) we shall have (by due ordering such Equation) 〈 math 〉: Or, RccH = 7RccA−14RqqAc + 7RqAqc−Aqqc.

Page  49XX. The Seven Roots of this Equation; are, so many streight Lines from some one Point of the Circumference, to the Seven Angles of an inscribed Regular Heptagon.

XXI. Of these Roots (putting H Affirmative,) the two least are Affirmative; the two next, are Negative; the two next to these, are again Affirmative; and, the greatest Negative.

XXII. And after the same manner we may proceed as far as we please: Collecting the consequent Multiplications and Sections, by the help of those Antecedent.

XXIII. And all such as are denominated by a Compound number (as 4 = 2 × 2, 6 = 2 × 3, 8 = 2 × 4 = 2 × 2 × 2, 9 = 3 × 3, &c.) may, with more convenience, (at lest, as to the Section, if not as to the Multiplication also,) be performed by two or more operations, according to the Components of such Compound number.

XXIV. But, both these, and those which are Denominated from Prime numbers, (as 3, 5, 7, 11, &c.) may (by such inscription of Quadrilaters) be Reduced to such Equations, as will contain as many Roots as is the number from which such Multiplication or Section takes its Denomination.

XXV. And, of these, those which are Denominated by an Even number, will afford Equations having Plain Roots; the Square Root of which Plains, are the subtenses of the Arches.

XXVI. But those which are Denominated by Odd numbers, afford Equations whose Roots are those subtenses.

XXVII. And, of these subtenses (as well in the one case as in the other,) the two least (which I look upon as the Principal Roots of the Equation,) are Affirmatives (supposing the Subtense of the Multiple Arch to be always put Affirmative;) the two next greater than these, Negatives; the two next Affirmatives; and so onward, Alternately, as long as there be Roots remain∣ing: save that, when the number is Odd, the greatest of all will be singular, whereas the rest go by Couples.

Page  50

CHAP. VI. Of the Proportion of the Base to the Legs of a Triangle, according as is the Angle at the Top of it.

I. THE noted Proposition of Pythagoras, (which is in Euclid, 47 è 1.) concerning the Square of the Base equal to the Squares of the two Legs containing a Right-angle: And two more in Euclid (Pr. 12, 13, è 2.) concerning the Excess, (in case the Angle at the Top be Obtuse;) or the Defect, (in case it be Acute;) of the Square of the Base, compared with the Squares of the two Legs: And some other Propositions in the foregoing Chapters, shewing what Proportion that Excess or Defect bears to a Rectangle of the Legs, in divers cases: Gave me occasion to pursue that Spe∣culation a little further; according to the following Propositions.

[illustration]

II. If by the Legs of a Triangle C, D, the Angle at the Top contained A, be a Right-angle, (or of 90 De∣grees;) and from thence a Perpendicular G, on the Base, cut this into two Segments χ, δ: The two Triangles hence arising, χ G C, G δ D, are like to the whole CDB; (because of one Angle common, the other a Right∣angle, and therefore the third equal to the third.) And therefore (the Triangles being here designed by their sides.)

〈 math 〉 and therefore 〈 math 〉 and therefore, Cq+Dq=(Bχ+Bδ=B: into χ+δ=B:=) Bq. That is,

The Square of the Base is equal to the two Squares of the Legs containing a Right-angle.

[illustration]

III. If the Angle A be 120 Deg. or 4/3 of a Right∣angle; and thence be drawn the Base two streight Lines G, Γ; making, with it, Angles equal to that at the Top: The Two Triangles χ GC, ΓδD, are like to the whole CDB, (because of one Angle common, another equal to A, and therefore the third equal to the third;) with a Triangle between GΓμ Equi∣later (because each of the Angles at the Base, and therefore that at the Top, are of 60 Degrees:) And the Base B=χ+δ+μ. And therefore, 〈 math 〉 and 〈 math 〉 Therefore, Cq+Dq+CD= (Bχ+Bδ+Bμ=B: into χ+δ+μ=) B:=Bq. That is,

The Square of the Base (of an Angle of 120 Degrees) is equal to the Squares of the Legs and a Rectangle of them.

Page  51IV. If the Angle A be 60 Degrees, or ⅔ of a Right∣angle; and from thence, G, Γ, making Angles with the Base equal to that of A: Then are, (for the same causes as before) χGC, ΓδD, like Triangles to CDB, (as before;) and GΓμ an Equilater Triangle com∣municating with them: And the Base, B=χ+δ−μ. (or χ−μ+δ.) And therefore,

[illustration]
[illustration]
〈 math 〉 and 〈 math 〉 And therefore, Cq+Dq−CD=(Bχ+Bδ−Bμ=B: into χ+δ−μ=B=) Bq. That is,

The Square of the Base (of an Angle of 60 Degrees) is equal to the Squares of the Legs, wanting the Rectangle of them.

(Note here, that, by χ, I understand the Base to the Legs CG; by δ, that of the Legs DΓ; by μ, that of GΓ; and, by B, that of CD, which is ever equal to χ+δ±μ, however these parts be intermingled. Which where it is +μ, is commonly more obvious to the Eye; but where it is −μ, is more perplex, and will need more consideration to discern; but it is equally true in both cases.)

V. If A be 135 Degrees (or 3/2 of a Quadrant,) and GΓ drawn (as before) to make the like Angle with the Base: The Triangles χGC, Γ δ D, will be like to CDB; and GΓμ will be Equi∣crural,

[illustration]
making the Angles at the Base, of 45 De∣grees, (so much as A wants of two Right-angles▪) and therefore, the Angle at the Vertex (which I shall call V,) of 90 Degrees. And therefore, (by § 2.) μq=2Gq, and 〈 math 〉 And the Base B=χ+δ+μ. And therefore, 〈 math 〉 and 〈 math 〉. Therefore, 〈 math 〉 into χ+δ+μ=B:=) Bq. That is,

The Square of the Base (of an Angle of 135 Degrees) is equal to the Squares of the Legs, with a Rectangle of them Multiplied into〈 math 〉

[illustration]

VI. If A be 45 Degrees: It will in like manner be shewed, that (because of B=χ+δ−μ.) 〈 math 〉 into χ+δ−μ=B:=) Bq. That is,

The Square of the Base (of an Angle of 45 Degrees) is equal to the Squares of the Legs, wanting a Rectangle of them Multiplied into〈 math 〉

Page  52VII. And, universally, what ever be the Angle A; it will (by like process) be shewed: That,

〈 math 〉. That is, The Square of the Base (whatever be the Angle at the Vertex) is equal to the Squares of the Legs, together with (if it be greater than a Right-angle) or wanting (if less than such) a Plain, which shall be, to the Rect-angle of the Legs, as a Portion in the Base-line, intercepted between two Lines from the Vertex, making at the Base a like Angle with that of the Vertex, to one of those two Lines so drawn.

VIII. Of this we are to give great variety of Examples in the following Chap∣ter, where this General Theorem is applied to particular cases: And which is further improved by these two ensuing Propositions.

[illustration]

IX. The Radius of a Circle, with the subtenses of two Arches, being given; the Subtense of their Aggregate is also given. For, supposing the subtenses of the given Arches to be A, E: The subtenses of their Remainders to a Semicircle, are also had: Suppose 〈 math 〉 And 〈 math 〉. And therefore, inscribing a Quadrilater whose opposite sides are A, ε; and E, α; one of the Diagonals is the Dia∣meter=2R; the other the Subtense of the Sum or Ag∣gregate of those Arches, suppose 〈 math 〉.

[illustration]

X. The same being given; the Subtense of the Difference of those Arches is also given. For, having (as before) A, α; E, ε; 2R: we have (by a Quadrilater duly inscribed) the Subtense of the Difference, 〈 math 〉.

XI. It is manifest also, (from what is before delivered,) that the same Tri∣angle GΓμ, doth indifferently serve for the Angle of 120 Degrees and of 60 Degrees: And, in like manner, for 135, and 45: And so, for any two Arches whereof one doth as much exceed as the other wants of a Quadrant. For, the Angle V is in both the same; and the Angles at the Base differ only in this: That, in one, the External Angle; in the other, the Internal, (which is the others Complement to two Right-angles;) is equal to the Angle of CD at the Vertex.

XII. Hence it follows: That, Of two Angles, where the Legs of the one are respectively equal to those of the other; the one as much exceeding a Right-angle, as the other wants of it: The Square of the Base in the one, doth as much exceed the two Squares of the Legs; as, in the other, it wants thereof.

[illustration]

XIII. And consequently, In any Right-lined Triangle, (however inclined,) the Squares of the Axis or Diameter, and of the half Bases twice taken; are equal to the Squares of the Legs. For, supposing C, C, the two halfs of the Base; and B, the Diameter or Axis of the Triangle, (meaning thereby a streight Line from the Vertex to the middle of the Base;) and B, β, the two Legs: It is manifest, that, of the two Angles at the Base (which are each others Complement to two Right-angles;) the one doth as much exceed, as the other wants of, a Right-angle: And there∣fore the Square of one of the Legs, as Bq, doth as much exceed; as the other, βq, doth come short of; Dq+Cq. And therefore, both together, Bq+βq=2Dq+2Cq.

Page  53XIV. And therefore, The Base, and Axis (or Diameter) of a Triangle remain∣ing the same; (however differently inclined:) the Aggregate of the Squares of the two Legs, remains the same.

XV. And the same is to be understood of the Squares of Tangents, of a Parabola, Hyperbola, Elipsis, (or other Curve Line having Diameter and Ordinates,) from the two ends of an Inscribed Ordinate, to the Point of the Diameter (produced if need be) wherein those Tangents meet.

XVI. The same may be likewise accommodated, to the Segments (of such Legs, or Tangents,) Cut off by Lines Parallel to the Base. Namely, The Squares of such Segments (intercepted by those Parallels) together taken, (the Axe of such Tra∣pezium remaining the same,) are the same: Whether such Trapezium be Erect, or however inclined. For such Segments, are still Proportional to their Wholes.

CHAP. VII. Application thereof to particular cases.

I. IF A be a Right-angle, (or of 90 Degrees,) GΓ are Co-incident, and μ=0. and therefore, 〈 math 〉. And consequently (by § 7, Chap. preced.) 〈 math 〉.

II. If A=120 Degrees; then is V (that is, the Angle contained of GΓ) =60 Degrees: (as being always the Difference of 2 A from two Right-angles:) And consequently GΓμ an Equilater Triangle, (for such also are the Angles at the Base; each of which is the Complement of A to two Right-angles:) And therefore, μ=G; and Bq=Cq+Dq+CD.

III. If A=60 Degrees: Then also is V=60 Degrees, and μ=G, as before. And therefore, Bq=Cq+Dq−CD.

IV. If A=135. Then V=90: And therefore, (by § 1.) μq=Gq+Γq, that is (because G=Γ,) μq=2Gq; and 〈 math 〉 And therefore, 〈 math 〉.

V. If A=45. Then also, V=90: And therefore, (as before) 〈 math 〉; and consequently, 〈 math 〉.

VI. If A=150 Then V=120. And therefore, (by § 2.) μq=Gq+Γq+GΓ; that is (because G=Γ,) μq=3Gq, and 〈 math 〉. And 〈 math 〉.

VII. If A=30 Then V=120. And therefore, (by § 2.) μq=Gq+Γq+GΓ; that is (because G=Γ,) μq=3Gq, and 〈 math 〉. And 〈 math 〉.

VIII. If A=157½ Then V=135. And 〈 math 〉 (by § 4.) And therefore, 〈 math 〉.

IX. If A=22½ Then V=135. And 〈 math 〉 (by § 4.) And therefore, 〈 math 〉.

X. If A=112½ Then V=45. And 〈 math 〉 (by § 5.) And therefore, 〈 math 〉.

XI. If A=6 − ½ Then V=45. And 〈 math 〉 (by § 5.) And therefore, 〈 math 〉.

Page  54XII. If A = 165 Then V = 150. And 〈 math 〉 (by § 6.) And there∣fore, 〈 math 〉.

XIII. If A = 15 Then V = 150. And 〈 math 〉 (by § 6.) And there∣fore, 〈 math 〉.

XIV. If A = 105 Then V = 30. And 〈 math 〉 (by § 7.) And there∣fore, 〈 math 〉.

XV. If A = 75 Then V = 30. And 〈 math 〉 (by § 7.) And there∣fore, 〈 math 〉.

XVI. If A = 172½ Then V = 165. And (by § 12.) 〈 math 〉. And therefore, 〈 math 〉.

XVII. If A = 7½ Then V = 165. And (by § 12.) 〈 math 〉. And therefore, 〈 math 〉.

XVIII. If A = 97½ Then V = 15. And (by § 13.) 〈 math 〉. And 〈 math 〉.

XIX. If A = 82½ Then V = 15. And (by § 13.) 〈 math 〉. And 〈 math 〉.

XX. If A = 142½ Then V = 105. And (by § 14.) 〈 math 〉. And 〈 math 〉.

XXI. If A = 37½ Then V = 105. And (by § 14.) 〈 math 〉. And 〈 math 〉.

XXII. If A = 127½ Then V = 75. And (by § 15.) 〈 math 〉. And 〈 math 〉.

XXIII. If A = 52½ Then V = 75. And (by § 15.) 〈 math 〉. And 〈 math 〉.

And, in like manner, we may proceed to lesser Arches, determined by quar∣ters of Degrees. For like as here, by help of § 4, 5, 12, 13, 14, 15. we have performed § 8, 9, 10, 11, 16, 17, 18, 19, 20, 21, 22, 23. which proceed to half Degrees: So by the help of these, we may proceed to Quarters of Degrees. And farther if we please: But I shall at present rest at half Degrees.

Moreover, assuming (as elsewhere proved) the Subtense of 36 Degrees, or the side of the inscribed Decagon; Namely, 〈 math 〉. (by 9 El. 13. and 4 El. 14. Or, 55, 56, Chap. 32.) we may, from thence, thus proceed.

XXIV. If A = 108 Then V = 36. And 〈 math 〉. And 〈 math 〉.

XXV. If A = 72 Then V = 36. And 〈 math 〉. And 〈 math 〉.

XXVI. If A = 144 Then V = 108. And μ = (by § 24.) 〈 math 〉. And 〈 math 〉.

XXVII. If A = 36 Then V = 108. And μ = (by § 24.) 〈 math 〉. And 〈 math 〉.

Page  55XXVIII. If A = 126 Then V = 72. And (by § 25.) 〈 math 〉. And 〈 math 〉.

XXIX. If A = 54 Then V = 72. And (by § 25.) 〈 math 〉. And 〈 math 〉.

XXX. If A = 144 Then V = 144. And (by § 26.) 〈 math 〉 And 〈 math 〉.

XXXI. If A = 18 Then V = 144. And (by § 26.) 〈 math 〉. And 〈 math 〉.

XXXII. If A = 153 Then V = 126. And (by § 28.) 〈 math 〉. And 〈 math 〉.

XXXIII. If A = 27 Then V = 126. And (by § 28.) 〈 math 〉. And 〈 math 〉.

XXXIV. If A = 117 Then V = 54. And (by § 29.) 〈 math 〉. And 〈 math 〉.

XXXV. If A = 63 Then V = 54. And (by § 29.) 〈 math 〉. And 〈 math 〉.

XXXVI. If A = 171 Then V = 162. And (by § 30.) 〈 math 〉. And 〈 math 〉.

XXXVII. If A = 9 Then V = 162. And (by § 30.) 〈 math 〉. And 〈 math 〉.

XXXVIII. If A = 99 Then V = 18. And (by § 31.) 〈 math 〉. And 〈 math 〉.

XXXIX. If A = 81 Then V = 18. And (by § 31.) 〈 math 〉. And 〈 math 〉.

XL. If A = 166½ Then V = 153. And (by § 32.) 〈 math 〉. And 〈 math 〉.

XLI. If A = 13½ Then V = 153. And (by § 32.) 〈 math 〉. And 〈 math 〉.

XLII. If A = 103½ Then V = 27. And (by § 33.) 〈 math 〉. And 〈 math 〉.

XLIII. If A = 76½ Then V = 27. And (by § 33.) 〈 math 〉. And 〈 math 〉.

XLIV. If A = 148½ Then V = 117. And consequently, (by § 34.) 〈 math 〉.

XLV. If A = 31½ Then V = 117. And consequently, (by § 34.) 〈 math 〉.

XLVI. If A = 121½ Then V = 63. And (by § 35.) 〈 math 〉.

XLVII. If A = 58½ Then V = 63. And (by § 35.) 〈 math 〉.

Page  56XLVIII. If A = 175½ Then V = 171. And (by § 36.) 〈 math 〉.

XLIX. If A = 4½ Then V = 171. And (by § 36.) 〈 math 〉.

L. If A = 94½ Then V = 9. And (by § 37.) 〈 math 〉.

LI. If A = 85½ Then V = 9. And (by § 37.) 〈 math 〉.

LII. If A = 139½ Then V = 99. And (by § 38.) 〈 math 〉.

LIII. If A = 40½ Then V = 99. And (by § 38.) 〈 math 〉.

LIV. If A = 130½ Then V = 81. And (by § 39.) 〈 math 〉.

LV. If A = 49½ Then V = 81. And (by § 39.) 〈 math 〉.

And, in like manner, by help of § 40, 41, &c. We may proceed to Arches determined by Quarters of Degrees; and further if need be.

Again, because the Subtense of 72 Degrees, is 〈 math 〉 and the Subtense of 60 Degrees is R: We may thence Collect the Subtense of their Difference, which is that of 12 Degrees; namely, R into 〈 math 〉; or 〈 math 〉. And thence proceed thus,

LVI. If A = 96 Then V = 12. And therefore, 〈 math 〉.

LVII. If A = 84 Then V = 12. And therefore, 〈 math 〉.

LVIII. If A = 138 Then V = 96. And (by § 56.) 〈 math 〉.

LIX. If A = 42 Then V = 96. And (by § 56.) 〈 math 〉.

LX. If A = 132 Then V = 84. And (by § 57.) 〈 math 〉.

LXI. If A = 48 Then V = 84. And (by § 57.) 〈 math 〉.

LXII. If A = 159 Then V = 138. And (by § 58.) 〈 math 〉

LXIII. If A = 21 Then V = 138. And (by § 58.) 〈 math 〉

LXIV. If A = 111 Then V = 42. And (by § 59.) 〈 math 〉

LXV. If A = 69 Then V = 42. And (by § 59.) 〈 math 〉

Page  57LXVI. If A = 156 Then V = 132. And (by § 60.) 〈 math 〉.

LXVII. If A = 24 Then V = 132. And (by § 60.) 〈 math 〉.

LXVIII. If A = 114 Then V = 48. And (by § 61.) 〈 math 〉.

LXIX. If A = 66 Then V = 48. And (by § 61.) 〈 math 〉.

LXX. If A = 169½ Then V = 159. And (by § 62.) 〈 math 〉.

LXXI. If A = 10½ Then V = 159. And (by § 62.) 〈 math 〉.

LXXII. If A = 100½ Then V = 21. And (by § 63.) 〈 math 〉.

LXXIII. If A = 79½ Then V = 21. And (by § 63.) 〈 math 〉.

LXXIV. If A = 145½ Then V = 111. And (by § 64.) 〈 math 〉.

LXXV. If A = 34½ Then V = 111. And (by § 64.) 〈 math 〉.

LXXVI. If A = 124½ Then V = 69. And (by § 65.) 〈 math 〉.

LXXVII. If A = 55 Then V = 69. And (by § 65.) 〈 math 〉.

LXXVIII. If A = 168 Then V = 156. And (by § 66.) 〈 math 〉.

LXXIX. If A = 12 Then V = 156. And (by § 66.) 〈 math 〉.

LXXX. If A = 102 Then V = 24. And (by § 67.) 〈 math 〉.

LXXXI. If A = 78 Then V = 24. And (by § 67.) 〈 math 〉.

LXXXII. If A = 147 Then V = 114. And (by § 68.) 〈 math 〉.

LXXXIII. If A = 33 Then V = 114. And (by § 68.) 〈 math 〉.

Page  58LXXXIV. If A = 123 Then V = 66. And (by § 69.) 〈 math 〉.

LXXXV. If A = 57 Then V = 66. And (by § 69.) 〈 math 〉.

LXXXVI. If A = 174 Then V = 168. And (by § 78.) 〈 math 〉.

LXXXVII. If A = 6 Then V = 168. And (by § 78.) 〈 math 〉.

LXXXVIII. If A = 141 Then V = 102. And (by § 80.) 〈 math 〉.

LXXXIX. If A = 39 Then V = 102. And (by § 80.) 〈 math 〉.

XC. If A = 129. Then V = 78. And (by § 81.) 〈 math 〉.

XCI. If A = 51 Then V = 78. And (by § 81.) 〈 math 〉.

XCII. If A = 163 ½ Then V = 147. And (by § 82.) 〈 math 〉.

XCIII. If A = 16 ½ Then V = 147. And (by § 82.) 〈 math 〉.

XCIV. If A = 106 ½ Then V = 33. And (by § 83.) 〈 math 〉.

XCV. If A = 73 ½ Then V = 33. And (by § 83.) 〈 math 〉.

XCVI. If A = 151 ½ Then V = 123. And (by § 84.) 〈 math 〉.

XCVII. If A = 28 ½ Then V = 123. And (by § 84.) 〈 math 〉.

XCVIII. If A = 118 ½ Then V = 57. And (by § 85.) 〈 math 〉.

XCIX. If A = 61 ½ Then V = 57. And (by § 85.) 〈 math 〉.

C. If A = 177 Then V = 174. And (by § 86.) 〈 math 〉.

CI. If A = 3 Then V = 174. And (by § 86.) 〈 math 〉.

CII. If A = 93 Then V = 6. And (by § 87.) 〈 math 〉.

CIII. If A = 87 Then V = 6. And (by § 87.) 〈 math 〉.

CIV. If A = 160 ½ Then V = 141. And (by § 88.) 〈 math 〉.

CV. If A = 19 ½ Then V = 141. And (by § 88.) 〈 math 〉.

Page  59CVI. If A = 109 ½ Then V = 39. And (by § 89.) 〈 math 〉.

CVII. If A = 70 ½ Then V = 39. And (by § 89.) 〈 math 〉.

CVIII. If A = 154 ½ Then V = 129. And (by § 90.) 〈 math 〉.

CIX. If A = 25 ½ Then V = 129. And (by § 90.) 〈 math 〉.

CX. If A = 115 ½ Then V = 51. And (by § 91.) 〈 math 〉.

CXI. If A = 64 ½ Then V = 51. And (by § 91.) 〈 math 〉.

CXII. If A = 178 ½ Then V = 177. And (by § 100.) 〈 math 〉.

CXIII. If A = 1 ½ Then V = 177. And (by § 100.) 〈 math 〉.

CXIV. If A = 91 ½ Then V = 3. And (by § 101.) 〈 math 〉.

CXV. If A = 88 ½ Then V = 3. And (by § 101.) 〈 math 〉.

CXVI. If A = 136 ½ Then V = 93. And (by § 102.) 〈 math 〉.

CVII. If A = 43 ½ Then V = 93. And (by § 102.) 〈 math 〉.

CXVIII. If A = 133 ½ Then V = 87. And (by § 107.) 〈 math 〉.

CXIX. If A = 46 ½ Then V = 87. And (by § 107.) 〈 math 〉.

And, in like manner, (by help of § 70, 71, &c. 92, 93, &c. 104, 105, &c. as was shewed at § 23.) we may proceed to Arches determined by Quarters of Degrees; or yet further, if there be occasion.

But we content our selves at present to rest at half Degrees. Having hereby sitted subtenses to every three halves of a Degree throughout the Semicircle.

Page  60

CHAP. VIII. Of the Canon of Subtenses, and Sines; Of Tangents also and of Secants.

FROM what is delivered in the foregoing Chapter; it is easie to con∣struct a Canon of Subtenses or Chords, in Surd Roots, to every Three∣halves of a Degree throughout the Semicircle. The halves of which Sub∣tenses, are the Right-sines for every Three-quarters of a Degree throughout the Quadrant.

(And thence, if need be, many Canons of Tangents and Secants, be deduced, in Surds Roots.)

And hereby, any who please, may either make new Tables, in Numbers, (to what accuracy he please,) or examin those already made.

For to every Subtense, to be successively sought, there will need but one extraction of the Square Root; (and, sometimes, not this;) the rest of the work being dispatched by only Addition and Subtraction; or, at most, Division also by 2 or 4.

As, for Example: Supposing the Radius of a Circle R = 1. Then (because these, in the same Circle be all equal,) C = D = 1. And likewise Cq = Dq = CD = 1. And B will be the Subtense of the Angle proposed.

Therefore, (by § 1.) the Square of the Subtense of 90 Degrees, Bq = Cq + Dq = 1 + 1 = 2. And the Subtense it self 〈 math 〉: Which is had by one extraction of the Square Root of the number 2; continued in Decimal Parts to to what accuracy we please. Suppose 〈 math 〉proximè.

Again, (by § 2.) the Square of the Subtense of 120 Degrees, is Bq = Cq + Dq + CD = 1 + 1 + 1 = 3. And the Subtense it self 〈 math 〉: Which is likewise had by one extraction of the Square Root of 3. Suppose 〈 math 〉proximè.

The Square of the Subtense of 60 Degrees, is (by § 3.) Bq = Cq + Dq − CD = 1 + 1 − 1 = 1. And therefore the Subtense B = 1.

The Square of the Subtense of 135 Degrees, is (by § 4.) 〈 math 〉. Which is had by adding 2, to the value of 〈 math 〉 already found at § 1. That is, 〈 math 〉. And so by one extraction of the Square Root of this number, we have the Subtense of 135 Degrees: Namely, 〈 math 〉proximè.

So (by § 5.) the Square of the Subtense of 45 Degrees, is 〈 math 〉. That is (by subtracting from 2 the value of 〈 math 〉 already found,) Bq = 0.58578643 ½ proximè. The Square Root of which, now to be extracted, is 〈 math 〉proximè.

Page  61And (by § 6.) the Square of the Subtense of 150 Degrees, is 〈 math 〉. Which is had, by adding 2 to the value of 〈 math 〉 all ready found. That is, 〈 math 〉. The Root of which (now to be extracted) is 〈 math 〉proximè.

Or thus: Because 〈 math 〉, (as will appear, either by the Squaring of this, or by extracting the Square Root of the Binomial 〈 math 〉) Having, as before, the value of 〈 math 〉; and (by one extraction now to be made) the value of 〈 math 〉 (or, it may be had by Multiplying the value of 〈 math 〉, by that of 〈 math 〉, already known; because 〈 math 〉) we have thence 〈 math 〉; and the half thereof 〈 math 〉. As before,

So (by § 7.) the Square of the Subtense of 30 Degrees, is 〈 math 〉, (which is had by Subduction only, the value of 〈 math 〉 being found before.) The Square Root of which (now to be extracted) is the Subtense 〈 math 〉proximè.

Or thus, (without extracting a Root;) because 〈 math 〉: Therefore, (the values of 〈 math 〉 and 〈 math 〉 being had before:) by Subduction only we have 〈 math 〉proximè; and (the half of this) 〈 math 〉proximè: As before,

And in the rest, (taking the Propositions or Paragraphs as they are before set down in the former Chapter,) there is need but of one extraction of the Square Root (and oft-times not of one,) for finding of each Subtense.

These Subtenses being thus had; the halves thereof are the Right-sines of the half Arch. As for Example.

Arches.Subtenses.Sines.Arches.
Degrees. 901.41421356 +0.70710678 +45 Degrees.
Degrees. 1201.73205081 −0.86602540 ½60 Degrees.
Degrees. 601.000000000.5000000030 Degrees.
Degrees. 1351.84775906 ½0.92387953 ¼67 ½ Degrees.
Degrees. 450.76536687 −0.38268343 ½22 ½ Degrees.
Degrees. 1501.93185165 +0.96592582 ½75 Degrees.
Degrees. 300.517638090.25881904 ½15 Degrees.

Now follows the Table of Subtenses in Surd Roots, answering to each three halves of a Degree throughout the whole Semicircle; (and consequently, of their Residuals to a whole Circle, whose Subtenses are the same with these:) Putting the Radius R = 1, and therefore C = D = 1, and likewise Cq = Dq = CD = 1: With references in the Margin to the Paragraphs of the for∣mer Chapter from whence they are derived.

Page  62〈 math 〉Page  63〈 math 〉Page  64〈 math 〉Page  65〈 math 〉Page  66 And, in like manner, we may proceed to design, by Surd Roots, the Sub∣tenses of Arches as small as we please, by a continual Bisection of these Arches. The halves of which Subtenses, are the Right Sines of the Half-Arches.

But, to design an intire Canon of Subtenses and Sines, answering to each single Degree, and the Sexagesims or first Minutes of such Degrees: Will (beside the extracting the Square Roots, of such Surds, in Numbers,) require also the Analysis (in Numbers) of Two Trisections, and of one Quinquisection of an Arch.

For, the former process reaching no farther than to the Subtense of 1½ De∣gree; and consequently to the sine of ¾ of a Degree, or of Min. 45 = 3 × 3 × 5: We may thence, by a Trisection twice performed; and a Quinquisection once, proceed to the sine of 1 Minute. But not by Bisections only, or operations thence deduced.

But, these operations being so (as is said) performed; the rest of the work is easily dispatched by help of § 9, 10. Chap. 6. for finding the Subtense of the Sum or Difference of those Arches whose Subtenses are already known.

CHAP. IX. Of Angles compared with they Arches on which the stand.

I. THAT The Angle of a Sector, is proportional to the Arch on which it doth insist; whether such Angle be at the Center, or at the Circumference: And, that such Angle at the Center, is double to that at the Circumfe∣rence: Is shewed by Euclide long since; and is generally known. But not so, in case such Angle be any where else, whether within or without the Circle. It will therefore be not amiss, to pursue that Notion a little far∣ther; as here followeth.

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II. If on the Diameter of a Circle D F, be for∣med, at the Center C, an Angle B A D; the inter∣cepted Arch B D, is Proportional to the Angle. (by 26, 27. El. 3. and 33, El. 6, of Euclid.) That is, the Arch intercepted B D, is such a Part of the whole Circumference; as is the Angle A, of Four Right-angles. And accordingly the Angle A, is said to be, of so many Degrees, as is the Arch B D.

III. If B A, D A, the Legs containing such Angle, after a Decussation at C, forming the Vertical Angle E, be continued, on the other side, to the Circum∣ference; the intercepted Arch G F, will, (by § 2.) be equal to B D, (because of the Vertical Angle E, equal to A:) And, consequently, the Aggregate of both B D + G F, is double to B D. That is, B D + G F = 2 B D.

IV. If at F the end of the Diameter, be formed a like Angle A; the inter∣cepted Arch H D, is likewise double to B D. That is, H D = B D + G F = 2 B D. Because (by 20. El. 3.) the Angle at the Center, is double to that at the Circumference, on the same Arch.

Page  67Or thus, because (by Construction) F H, G C B, are Parallels: (as making equal Angles with F D:) Therefore, H B = F G = B D. And, consequently, H D = H B + B D = F G + B D = 2 B D. (By § 3.)

V. If at K, (any other Point of the Diameter within the Circle,) such Angle A be made, (with its Vertical E:) the Aggregate of the two Arches intercep∣ted, F L + M D, is Double to B D. That is, F L + M D = F G + B D = 2 B D. For, (drawing the streight Line A M,) the two Internal Angles F M L + M F D, are equal to the External M K D = A = H F D (which is an Angle at the Circumference.) And therefore, the Arches opposite to those F L + M D, equal to the Arch opposite to this H D = 2 B D. (By § 4.)

Or thus, because F H, L M, be Parallels (as making like Angles with F D,) therefore, F L = H M, and F L + M D = H M + M D = H D = 2 B D. (By § 4.)

VI. If at N, a Point of the Diameter produced, without the Circle; be formed a like Angle A; the Difference of the two intercepted Arches, Q D − P F, is equal to the same H D, or the Double of B D. For, P Q, F H, being Parallels, (as making like Angles with D F produced,) and therefore, P F = Q H: Therefore, Q D − P F = Q D − Q H = H D = 2 B D. (By § 4.)

VII. The same will hold, though neither of the Legs containing the Angle do pass through the Center, (and therefore lie not upon a Diameter;) as I shall now shew in the several Cases.

VIII. If M L, μ λ, make M K μ an Angle at K any where, within the Circle, Let B c β be a like Angle at the Center, and the Legs of this Parallel to those of that. And, by the Angular Point K, draw the Diameter F K D. Then is (by § 5.) M D + L F = 2 B D: And μ D + λ F = 2 β D. Therefore, (the Sum or Difference) M μ (= M D ± μ D) + L λ (= L F ± λ F) = (2 B D ± 2 β D =) 2 B β.

[illustration]

IX. In like manner: If H F η be an Angle at the Circumference: And, at the Center, B C β like to it, and with Legs Parallel to those: And F D a Di∣ameter. Then (by § 4.) H D = 2 B D, and η D = 2 β D. Therefore, H η (= H D ± η D) = (2 B D ± 2 β D =) 2 B β.

[illustration]

Page  68X. The same is to be understood, in case one of the Legs touch the Circum∣ference at F, (the Points F, H, being in this case Co-incident; the Arch F H vanishing to nothing, and the Arch intercepted H η, the same with F η.) For here also, H F D = 2 B D, and η F D = 2 β D; and therefore, H F η (= H F D ± η F D) = (2 B D ± 2 β D =) 2 B β.

[illustration]

XI. In like manner: If Q N χ be an Angle without the Circle, whose Legs cut it in P, π: And, at the Center, a like Angle and like sited B C: And N F D a Diameter produced. Then (by § 6.) Q D − P F = 2 B D, and χ D − π F = 2 β D. And therefore Q D ± χ D wanting P F ± π F, that is Q χ − P π, is equal to 2 B D ± 2 β D = 2 B β.

[illustration]

XII. The same is to be understood, in case one or both of the Legs do not cut, but only touch the Circle. For then the Points P, Q; or π, χ; (or both,) being Co-incident; the rest proceeds as before. For still Q D − P F = 2 B D, and χ D − π F = 2 β D; and therefore, (the Sum or Difference) Q χ − P π = 2 B β.

[illustration]

XIII. But in case one or both of the Legs pass by the Circle, and neither cut, nor so much as touch it: It doth not concern the present business; for such Angle doth not insist on a Circular Arch. The whole therefore (thus demonstrated) amounts to this General.

XIV. If a Circle be cut (or at least touched) by two streight Lines, making an Angle: (and so, when continued, intersecting each other:) The Sum (if their intersection be within the Circle,) or Difference (if without,) of the two Arches intercepted by them (produced, if need be,) or (if their intersection be at the Cir∣cumference) the single Arch by them intercepted; is Double to the Arch of a like Angle at the Center.

FINIS.
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