unless the ambiguity be removed by one of the following rules: 1. In any right-angled spherical triangle, an oblique angle and its opposite side are always of the same species. For, by (205), sin b cot A tan p, in which, since sin b is always positive, cot A and tan p must always have the same sign, that is, A and p must be of the same species. 2. When the two sides about the right angle are of the same species, the hypothenuse is less than 90°, but when they are of different species, the hypothenuse is greater than 90°. in which, if cos p and cos b have the same signs, cos h will be positive, but if they have unlike signs, cos h will be negative. 171. In the solution of right-angled spherical triangles, there will be six cases to consider, in which there may be given, respectively, I. The hypothenuse and an oblique angle. II. The hypothenuse and one side. III. One side and its adjacent oblique angle. CASE I. 172. Given the hypothenuse and an oblique angle. Let there be given in the rightangled spherical triangle ABC, the hypothenuse h and the oblique angle A; to solve the triangle. To find p. Make p the middle part, and we have, by Napier's rules, or by (200), B h C or, by logarithms, log sin plog sin A+ log sin h. (208) To find b. Make the complement of A the middle part, and we have, by Napier's rules, or by (202), To find B. Make the complement of h the middle part, and we have, by Napier's rules, or by (199), Thus, b and B, by observing the algebraic signs, are determined without ambiguity; and p, though determined by its sine, is not ambiguous, since it must be of the same species as A (Art. 170). EXAMPLES. 1. Given in a right-angled spherical triangle ABC, rightangled at C, the hypothenuse h equal to 105° 34′, and the angle A equal to 80° 40′; to solve the triangle. Solution. By (208), By (210), h, log sin+9.983770 log tan-10.555053 4, log sin+9.994212 log cos+ 9.209992 By (212), log cos- 9.428717 log tan+10.784220 p, log sin+9.977982 b, log tan- 9.765045 B, log cot-10.212937 Hence, p=71° 54′ 33′′, b=149° 47′ 37′′, B=148° 30′ 54′′. 2. Given in the spherical triangle A B C, right-angled at C, the hypothenuse h equal to 70° 23' 42", and the angle A equal to 66° 20' 40"; to find the other parts. Ans. P, 59° 38′ 26′′; b, 48° 24′ 15′′; B, 52° 32′ 55′′. CASE II. 173. Given the hypothenuse and one side. Let there be given (Fig. Art. 172) the hypothenuse h and the side p; to solve the triangle. To find A. Make p the middle part, and we have, by Napier's rules, or by (200), To find B. Make the complement of B the middle part, and we have, by Napier's rules, or by (203), or, by logarithms, cos Bcoth tan p, log cos Blog cot h + log tan p. (215) To find b. Make the complement of h the middle part, and we have, by Napier's rules, or by (198), Here, as in the preceding article, b and B are determined without ambiguity, for there is only one angle less than 180° corresponding to a given cosine; and A must be of the same species as p. EXAMPLES. 1. Given in a right-angled spherical triangle A B C, the hypothenuse h equal to 91° 42', and the side p equal to 95° 22' 30"; to solve the triangle. Ans. A, 95° 6'; B, 71° 36′ 45′′; b, 71° 32′ 12′′. 2. Given in a right-angled spherical triangle, the hypothenuse equal to 70° 23′ and a side equal to 48° 24'; to solve the triangle. CASE III. 174. Given one side and its adjacent oblique angle. Let there be given (Fig. Art. 172) the side and the angle A; to solve the triangle. To find B. Make the complement of B the middle part, and we have, by Napier's rules, or by (207), or, by logarithms, cos B sin A cos b, log cos Blog sin A+ log cos b. (218) To find p. Make b the middle part, and we have, by Napier's rules, or by (205), To find h. Make the complement of A the middle part, and we have, by Napier's rules, or by (202), EXAMPLES. 1. Given in a spherical triangle ABC, right-angled at C, the side b equal to 29° 46′ 8′′, and the angle A equal to 137° 24′ 21′′; to solve the triangle. Ans. B, 54° 1′ 16′′; p, 155° 27′ 54′′; h, 142° 9′ 13′′. 2. Given in a spherical triangle ABC, right-angled at C, the side p equal to 149° 47' 23", and the angle B equal to 80° 40'; to find the other parts. CASE IV. 175. Given one side and its opposite oblique angle. Let there be given in a spherical triangle ABC, right-angled at C, the side p and the opposite angle A; to solve the triangle. To find h. Make p the middle part, and we have, by Napier's rules, B A b b C or by (200), To find b. Make b the middle part, and we have, by Napier's rules, or by (205), or, by logarithms, sin b cot A tan p, log sin b = log cot A+ log tan p. (225) To find B. Make the complement of A the middle part, and we have, by Napier's rules, or by (206), |