### 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 R^{6}; 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.