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13. In any right angled triangle, the hypothenuse is to one of the legs, as the radius to the sine of the angle opposite to that leg; and one of the legs is to the other, * the radius to the tangent of the angle opposite to the atter.
Let ABC be a triangle, right-angled at C B, and let AR on the leg AB, be the radius of the tables. With centre A and T radius AR, describe an arc to cut the hy- D pothenuse in D, and draw DH, TR, perpendicular to AB. Then DH is the sine, TR the tangent, and AT the secant, of the ATH R B arc Dr, or angle A: and the similar triangles ABC, ART, AHD, give
Ac: ch:: AD : DH :: rad: sin A; and AB : BC :: AR : RT :: rad: tan A.
Cor. To the hypothenuse as a radius, each leg is the sine of its opposite angle; and to one of the legs as a radius, the other leg is the tangent of its opposite angle, and the hypothenuse is the secant of the same angle.
14. In any plane triangle, as one of the sides, is to another, so is the sine of the angle opposite to the former, to the sine of the angle opposite to the latter. , Let ABC be a plane triangle, C in which it is required to determine the relation which subsists between the sides Ac and Bc. ? With the angular point A as a dif A ā HT&TR
centre, and the distance Ac, equal to the radius of the tables, describe the semicircle Gech. From c draw có parallel to CB, and cd perpendicular to AB : draw also eA parallel to cB, and from the point of intersection e (of that line with the circle) demit the perpendicular ef on BA produced. Then cd is the sine of the angle cAB, to the radius Ac, and ef is the sine of the angle eAG = angle CBA, to the same radius. But Ac: Bc :: Ac: be, because of the parallels Bc and be, :: Ae: be, because Ae = Ac, :: €f: cd, because of the similar triangles Aff, bed. That is, Ac: BC :: sin B : sin A. In a similar manner it may be shewn that Ac: AB :: sin B : sin C, and AB : Bc :: sin c : sin A. And by drawing a figure for each case, it will be seen that the circumstance of any one of the angles being obtuse will make no difference in the demonstration.
Otherwise. From c the vertical angle of the C triangle let fall the perpendicular cD upon AB, or AB produced, according as the angles A and B are both acute, g or one obtuse. - ATD
15. In any plane triangle it will be, as the sum of the sides about the vertical angle, is to their difference, so is the tangent of half the sum of the angles at the base, to the tangent of half their difference. By the preceding prop. Ac: Bc :: sin B : sin A, , , ... by comp, and div. Ac,+ Bc. Ac-Bc: sin B + sin A : Sin B - Slil A.
16. In any plane triangle it will be, as the base, to the sum of the two other sides, so is the difference of those sides, to the difference of the segments of the base made by a perpendicular let fall from the vertical angle.
From the centre c (namely, the vertical angle of the triangle ABc) with the distance of the greater side AC describe the circumference of a circle, meeting AB, cB, produced, in the points E, F, G. Then, it is obvious that GB is equal to the sum of the sides, AC, CB, and FB equal to their difference. And because co is perpendicular to AE, AD is equal to DE, (Euc. iii. 3). Wherefore, of the two AB, BE, one is the sum of the seg- 27 ments of the base DA, DB, and the AN E other their difference. But, (Euc. iii. 35) the rectangle under cB, and F FF, is equal to the rectangle under AB and B.E. Consequently (Euc. vi. 16) AB : GB :: BF : BE; which agrees with the proposition.
PROP, XVII. 17. In any plane triangle it will be, as twice the rectangle under any two sides, is to the difference of the sum of the squares of those two sides and the square of the base, so is the radius to the cosine of the angle contained by the two sides. Let ABC be a plane triangle. From A, C one extremity of the base AB, draw AD per- p endicular to the opposite side Bc : then Euc. ii. 12, 13) the difference of the sum of the squares of Ac, cB, and the square of * B the base AB, is equal to twice the rect- D
angle Bc.co. But twice the rectangle C BC. CA, is to twice the rectangle Bc. CD, s’s that is, to the difference of the sum of the
squares of Ac, BC, and the square of AB, as ca to cD; that is, (prop. 13) as radius to the sine of cAD, or, as radius to the cosine of Ach. Cor. When unity is assumed as radius, we have Ac2 + BC” – AB2 COSC E 2CB. cA
18. As the sum of the tangents of any two unequal angles, is to their difference, so is the sine of the sum of those angles, to the sine of their difference. Let BCA, ACD, be the two proposed C angles, BA and AD their tangents to the radius CA. Take IA = AD, join c1, and draw DE, de, Ix, ik, perpendicular to Bc. Then, it is evident, since 1A = AD, and CA perpendicular to BD, that c1 = cD, 1CA = DcA, and therefore BCI = BcA — AcD: also that ed is the sine of the sum of the given angles, and ik the sine of their differe ce,to the assumed radius Ac. Now, by reason of the similar triangles BDE, BIK, it will be, BD (= BA + AD): B1 (= BA – AD):: DE: IR. Also, because the triangles CDE, Cde, are similar, as well as the triangles CIK, cik; we have - CD: cal:; DE: de, - and cI (= CD): Ci (= ca) :: IK : ik; therefore DE: Ik:: de : il. And consequently, BB : B1:: de : ik; - that is, tan BCA + tan DCA: tan BCA – tan DCA :: sin (BCA + DCA): sin (BCA - DCA). Cor. Hence it follows, that the base of a plane triangle BCD, is to the difference of its two segments B.A.,
AD, as the sine of the whole angle at the vertex, to the sine of the difference of the angles at the base.
19. As the sine of the difference of any two unequal angles, is to the difference of their sines, so is the sum of those sines, to the sine of the sum of the angles. Let A and B be the two unequal an- C gles; both being bounded on one side by the line AB, and on the other by lines, which, produced, meet at c, and form a rectilinear triangle AcB. Demit cD perpendicularly from c upon AB; make Da A D a B = DA, and join ca. Then we shall have a B = DB – DA, Ca = CA, cad = CAD, cab = supp. CAB; and conseq. (ch. i. 19 c) sin cAB = sin cab. Also (by the o sin c = sin (A + B); and agb = Aac – B = A — B. Hence, in A AcB, sinc, or sin (A + B): sin A :: AB : cB. Also, in A acB, sin C, or sin (A – B): sin A :: a B : CB. Therefore, ab: AB :: sin (A – B); sin (A + B). Again (prop. 14) CB : CA :: sin A : sin B. ...comp. and div, co-cA: CB + cA : sin A — sin B : sin a + sin B. But, by the inversion of prop. 16, a B : CB - CA :: CB + CA: AB. Therefore, comparing the corresponding antecedents and consequents, in the 3d and 5th analogies, there results, sin (A – B): sin A — sin B:: sin A + sin B : sin (A + B). Cor. If A and B be to each other as n + 1 to n, then the last proportion will become sin A : sin (n + 1) A — sin na :: sin (n + 1) A + sin nA : sin (2n + 1) A.
20. Of the propositions in this chapter, some are useful in the solutions of plane triangles, and will find their applications in the next chapter; others are useful