Second Case: given A, B, C. -cos S. cos(S-4). cos(S-B). cos(S-C)\ ́· B) log Sin a=log 2 + logr-log Sin B — log sin C + } { log( — Cos S) + log Cos (S— A) + log Cos (S +log Cos (S-C)}. + { log( − Cos S) + log Cos (S-A)-log Sin Blog Sin C}. + {log Cos (S—B) + log Cos (S— C) — log Sin Blog Sin C}. log Tanalogr + { log( — Cos S) + log Cos (S A) -log Cos (S-B) - Cos (S-C)}. [5] Assume cos =cos B.cos C; log Cos = log Cos B + log Cos C-logr: then log Cos a = logr + log 2 + log Cos (4+0) + log Cos(4-0). The same observations as in the analagous cases of plane triangles will apply to the numerical accuracy of the different methods in these two cases. Third Case: given a, b, C. To determine A and B: log Tan (A+B)=log Cos(a−b)+log Cot & C―log Cos{(a+b). log Tan (4-B) = log Sin (a−b) + log Cot & C-log Sin (a+b). A = {(A + B) + (A — B), B={(A+B) — }; (A — B). [2] Assume tan a = tan a. cos C; log Tanalog Tan a + log Cos C — logr: and let B-b-a, then log Tan A=log Tan C+ log Sin a- log Sin B. To determine c. [1] 2 (sin c) vers (a - b) + sin a. sin b.vers C. log Tan 0={log Sina + log Sin b + log Vers C-log Vers (a — b) } . then 2 (sin c) vers (a - b). (sec )2; = log Sin clog Sec 0+{log Vers (a - b) - logr-log 2}. [2] (sinc)2 = (sin 1⁄2. a + b)2 — sin a. sin b (cos / C)2. Assume (sin 0)2= sin a. sin b. (cos C)° log Sin = {log Sina + log Sin b} + log Cos C-logr: then (sin c)2= sin (§. a + b + 0). sin (. a + b − 0) ; log Sinc={log Sin (.a+b+0) + log Sin (. a + b−0)}· [3] Assume tan a = cos C. tan a; log Cos c=log Cos a + log Cos B-log Cos a. If c is small, this formula is not sufficiently accurate. log Sin c = log Sin C + log Sin a — log Sin A. Fourth Case: given A, B, c. To determine a and b : log Tan (a+b)=log Cos (4-B) + log Tanic-log Cos(4-B). tan (a - b) = log Tan (a-b) = log a = (a + b) + Sin (A–B) — log Sin (A+B) + log Tanc. log Tan.alog Tan c + log Cos a-log Cos B. log Sin C = log Sin A + log Sin c-log Sin a. [2] 2 (sin C)2 = 2(cos }. A + B)2 + vers c.sin A.sin B. log Tan 0={log Vers c + log Sin A + log Sin B + log r — log 2} -log Cos(A+B): then sin C = cos (A + B). sec 0 ; log Sin Clog Cos (A + B) + log Sec — log r. [3] Assume cot a=tan A.cos c ; log Cot a = log Tan A+ log Cos c-logr: and let ẞ=B-a, then log Cos C = log Cos A+ log Sin ẞ - log Sin a. This is not sufficiently accurate, if C is small. Fifth Case: given a, b, A. To determine B: log Sin Blog Sin A + log Sin b―log Sin a. sin A, B=90°, when A and a are of the same species; when A and a are of different species, there is no solution. If > sin A, and < 8, there are two supplemental values sin a sin b of B, when A and a are of the same species; when of different species, there is no solution. 1, that value only of B is admissible, which is of log Cot & Clog Tan (A+B) +log Cos (a + b) — log Cos(a - b) log Cosẞ=log Cos a + log Tan b— log Tan a: then Ca±ẞ; + or -, according as a and ẞ are of the same or different species, if sin a sin b sin a > 1: if <1, and > sin A, then C sin b log Sin (C + 0) = log Tan b + log Sin - log Tan a. log Sin c= log Sin a + log Sin C — log Sin A. log Tanc-log Tan (a+b) +log Cos (A+B)-log Cos(4-B). |