CHAP. X. Formule of Solution for Right-angled Spherical Triangles. Affections of Sides and Angles. Circular Parts.-Naper's Rules. - Quadrantal Triangles. Examples. PROP. XVII. PROBLEM. It is required to investigate formulæ of solution for all the cases of right-angled spherical triangles *. These are to be derived from the fundamental expressions [a], [b], [c], given in page 139: let C be the right angle and = 90°; ... cos. C = 0; spectively, for cos. a and cos. b, in the expressions for cos. A, cos. B, and we have * Spherical triangles that have one right angle only, are the subject of investigation: and spherical triangles that have two right angles, and three right angles, are excluded. Now in [1] a, b, c are involved, and two of these being given the third may be found: in [2] A, c, b are involved: and if we choose, symbolically, to represent by (A, c, b) the form in which three quantities as A, c, b, are involved, then, similarly, the other several forms that can arise by combining angles and sides may be thus represented : (c, a, b); { (A, c, b), (B, c, a) }; { (B, c, b) (A, c, a) } { (A, a, b) (B, a, b) }; { (A, B, a) (A, B, b) }; (A, B, c) : which are in number 10, as they ought to be; the combination 5.4.3 in 5 things, 3 and 3 together, being = 10: those combinations that are similar, such as the second and third, are included within brackets. The forms [1], [2], [3], have been already deduced, and the remaining ones must be deduced from them, and from the form sin. A sin. B , by the common process of substitution and elimination, that is, cos. c = cos. a.cos. b.[1] sin. a = sin. b (c, a, b), (A, c, b) (B, c, a) cos. B cot.c.tan.a......... [2] ..[3] .. sin. B , or sin. asin.c.sin. A [5] sin. c (A, a, b): cos. A=cot. c. tan. b by [2], but cot. c = and cos.c=cos. a.cos.bby [1], and sin. c cos.a.cos. b. sin. A cos. A tan. b. (B, a, b) is similar, and , sin. a sin. by [5], p. 144, .. cot. A = cot. a. sin. b......[6] .. cot. B = cot. b.sina....[7] (A, B, a): divide [2] by [4], then Cos B = COS. C sin. cos. b from [1] or, cos. A = cos.a. sin. B......[8] (A, B, b) is similar, and cos. B = cos. b. sin. A. [9] (A, B, c): multiply together [6] and [7] and cot. A.cot. B = cos.b.cos.a=cos.c... ...[10] ... ....... Thus are all the forms easily deduced from [1], [2], [3]. Indeed, the quantities that are to be combined, indicate plainly the proper process of elimination; but these forms, although very simple and fitted for logarithmic computation, are not easily remembered, and therefore an artificial memory has been supplied to the Student and Computist, by rules, known by the title of Naper's Rules for Circular Parts; and in the whole compass of mathematical science there cannot be found, perhaps, rules which more completely attain that, which is the proper object of rules, namely, facility and brevity of computation. The rules and their description are as follow: Description of Circular Parts. The right angle is not considered; the complements of the two other angles, the complement of the hypothenuse, the two sides, making in all five quantities, are called by Naper, circular parts. Any one of these circular parts may be called a middle part (M), and, then, the two circular parts immediately adjacent to the right and left of M are called adjacent parts; the other two remaining circular parts, each separated from M the middle part, T by an adjacent part, are called opposite parts, or opposite extremes : thus, let the side a be M, then, comp. B, or 90° - B, and b are adjacent parts, If 90° - A be M, 90°-c and b are adjacent parts; 90°-B and a, opposite parts. This necessary explanation being premised, we come to Naper's Rules. 1. The rectangle of the sin. M and radius = rectangle of the tangents of adjacent parts. 2. The rectangle of the sin. M and radius = rectangle of the cosines of the opposite parts. There is no separate and independent proof of these rules, but the rules will be manifestly just, if it can be shewn that they comprehend every one of the ten results, [1], [2], [3], &c. given, in pages 144, 145; for, those results solve every case in rightangled spherical triangles. Μ. Adjacent Parts. Opposite Parts. 2. cos. A=cot. c. tan. b, or, sin. (90°-A)=tan. (90° −c) tan. b. 3. cos. B=cot. c. tan. a, or, sin. (90°-B)=tan. (90°-c) tan. a...... = A, or, { 1. cos.c=cos.a. cos. b, or, sin. (90°-c)=cos. a. cos. b 4. sin. b = sin.c.sin. B, or, sin. b=cos. (90° - c) cos. (90° - B). 5. sin. sin.c.sin. sin. a=cos. (90°-c) cos. (90° 6. cot. A = cot. a. sin. b, or, sin. b=tan. (90°-A).tan.a... 17. { cot. B = cot. b. sin. a, or, sin. a=tan. (90°-B). tan.b.. 8. cos. A =cos. a. sin. B, or, sin. (90°-A)=cos. a. cos. (90° - B). 9. cos. B=cos. b. sin. A, or, sin. (90°- B)=cos. b. cos. (90°-A). 10. cot. A. cot. B=cos. c, or, sin. (90° - c) =tan.(90°-A) tan. (96° - B) 90° - 90°-A, 90°- B 90° a, b .... ..... b .... a 90°, 90°-B b 90° - A, a a 90° B, b |