Napier's Analogies. they are both adjacent is to the tangent of half the difference of the other two sides; that is, in the spherical triangle ABC (figs. 32 and 33), sin.(A+C): sin.(A–C)=tang.36: tang (a—c). (350) Proof. The quotient of (343), divided by (345) is, by an = (352) sin. (AC) sin. (sc) sin. (s—a)' If we make in equation (47) we have This equation, substituted in the second member of (352), 80. Theorem. The cosine of half the sum of two angles of a spherical triangle is to the cosine of half Napier's Analogies. their difference, as the tangent of half the side to which they are both adjacent is to the tangent of half the sum of the other two sides; that is, in the spherical triangle ABC (figs. 32 and 33), cos. (A+C): cos.(A–C)=tang. b: tang.(a+c). (354) Proof. The product of (343) and (345) is, by a simple reduction, sin. s - sin. (s. b) = (355) ·6) cos. (AC) sin. s + sin. (s If in equation (47) inverted we make This equation, substituted in (355), gives Napier's Analogies. 81. Scholium. In using (350) and (354), the signs of the terms must be attended to by means of Pl. Trig. § 61. 82. EXAMPLEs. 1. Given in a spherical triangle two angles 158°, and = 98°, and the included side = 144°; to find the other sides. 2. Given in a spherical triangle two angles 170°, and= 2o, and the included side = 92°; to find the other sides, Ans. a 103° 6' 44", c = 11° 17′ 16′′. Three angles given. 83. Problem. To solve a spherical triangle, when its three angles are given. Solution. If A, B, C are the angles of the given triangle, and a, b, c its sides, 180° - A, 180° — B, 180° — C are the sides of the polar triangle, and 180°-a, 180°-b, 180° -c the angles of the polar triangle, the sides are then given in the polar triangle; to find the angles. For this purpose we may use the formulas of the preceding problem. 84. Corollary. Applying (331) to the polar triangle gives COS. C = cos. C+cos. A and B sin. A sin. B (357) 85. Corollary. Equations (335-337) give, for the polar triangle, if we put 86. Corollary. Equations (340-342), applied to the polar triangle, give cos, a = √(cos. (S—B) cos. (S-C) sin. B sin. C (362) cos. (S-A) cos. (S-C) (363) sin. A sin. C 87. Corollary. Equations (343–345), applied to the polar 88. Corollary. Equation (332), applied to the polar tri angle, is 2 (sin. c)2 = cos. C cos. (A+B) sin. A sin. B which may be used like equation (338). (368) 89. EXAMPLE. Given in the spherical triangle ABC, the three angles equal to 89o, 5o, and 88°; to solve the triangle. Ans. The three sides are 53° 10′, 4°, and 53° 8'. 90. Theorem. The sine of half the sum of two sides of a spherical triangle is to the sine of half their difference, as the cotangent of half the included angle is to the tangent of half the difference of the other two angles, that is, in ABC (figs. 32 and 33), sin.(a+c): sin. (a–c)=cotan. 2 B: tang. 2 (A-C). (369) 12 |