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.