(744) (745) (746) (747) (748) (750) This equation is the same as the proportion cos. M cos. N: sin. M sin. N:: y: x; hence, by the theory of proportions, cos. M cos. N sin. M sin. N: ços. M cos. N + sin. M sin. Ny-x:x+y, or, by (104) and (116), cos. (M+N): cos. (MN)::y-x:y + x; which may be written as in (744). 60. Theorem. The sine of half the sum of two angles of a spherical triangle is to the sine of half their difference, as the tangent of half the side to (749) which 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. 4. and 5.), sin. (AC) sin. † (A — Demonstration. The quotient of (702), divided by (704) is, by an easy reduction, A B P sin. (s tang. A sin. Hence, by (736) and (737), we have sin. + (A + C) sin. (sc) sin. (sa) sin. (AC) sin. (s S- c) — sin. (s 1 sin. A' + sin. B' sin. A' sin. B' = = If we make in this equation SA' + B'b, and (753) becomes sin. (s sin. (s But we have by (228), accenting the letters so as not to confound them with the angles of the triangle, A'B' a— c; c) + sin. (s c) — sin. S- s) tang. tang. sin. (A + C) which is the same as (750). a) a) c) a) = cos. (A + C): cos. † (A (A' + B') tang.b This equation, substituted in the second member of (752), gives tang. b = tang. (a a) c) (a+c). i C) :: tang, b (751) (752) (753) (754) 61. Theorem. The cosine of half the sum of two angles of a spherical triangle is to the cosine of half their difference, as the tangent of half the side to (758) 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. 4. and 5.) (755) (756) (757) (759) (760) (761) (762) (763) (764) (765) (766) Demonstration. The product of (702) and (704), is, by a simple reduction, tang. A tang. C= Hence, by (743) and (744), sin. (s—b) sin. s tang. (A'B') tang. sin. s sin. (s cos. (A + C) In using (749) and (758), the signs 62. Scholium. (767) of the terms must be attended to by means of (496). 1. Given in the spherical triangle ABC (figs. 4. and 5.) to find sides a and c. A = 158°, C' ≥ (A + C) b (a — c) Solution. By (750), 128°. sin. (ar. co.) 10.10347 = 30°. sin. 9.69897 72°. tang. 0.48822 62° 53', tang. 0.29066 By (759), (A + C) ¿ (A — C) 1 b = EXAMPLES. to find a and c. = = = 98°, 128°. cos, (ar. co.) 10.21066 n. = 30°, COS. 9.93753 72°. tang. 0.48822 (a+c) 103°, b = 144°; tang. Ans. A 170°, C 2°, b = 92°; = a = 165° 53', c = 40° 7'. 0.63641 n. 2. Given in the spherical triangle ABC (figs. 4. and 5.) Ans. a = 103° 7', 63. Theorem. The greater side of a spherical triangle is opposite the greater angle. (768) (769) Demonstration. The first and third terms of the proportion (750), · † : tang. (a — c), are, by (496), both positive, since (770) than 180° and b is less than 90°. fourth terms must then be both positive or both negative at the same time. But as (AC) and (771) 1⁄2 (a c) are both less than 90°, these terms can be negative only when A is less than C and a less than c which agrees with (768). (774) (A + C) is less The second and 64. Theorem. Of two sides of a spherical triangle the one which differs most from 90° must be (772) opposite the angle which differs most from 90°; and this side and angle must be both greater or both less than 90°. Demonstration. Let the side a differ more from (773) 90°, than does the side b; then by (507) sin. a is less than sin. b. But, by (598), sin. a sin. b:: sin. A: sin. B. Hence, sin. A is less than sin. B, and by (507) the (775) angle A differs more from 90° than does the angle B which agrees with the first part of (772). Again, if a is also greater than b it must be greater than 90°; and the opposite angle A must, by (768), (776) be greater than the angle B, and, by (775), differing more from 90° must also be greater than 90°. But if a is less than b, it must, by (773), be acute; and A (777) must, by (768), be less than B, and, by (775), it must also be acute; which is the second part of (772). |