A treatise of angular sections by John Wallis ...
Wallis, John, 1616-1703., Wallis, John, 1616-1703. Treatise of algebra.
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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.