A treatise on spherical trigonometry, and its application to geodesy and astronomy, with numerous examples. By John Casey.

58 Solution ofSpherical Triangles.
Second Pair of Cases.
63. Given two sides a, b, and the angle A opposite to one of
them, to calculate the remaining parts.
The sought parts are found by the following equations:
sin A. sin b
sin B== (257)
sin a
tan = tan (a - b) si( ) (258)
sin (A -B)'
tan x C= cot (A-B) sin (a - b) (259)
sin (a + b)'
The formula (257) gives for B two values BI, 180~ - B1, if
sin A sin b be less than sin a. In order that either of these
may be admissible, it is necessary and sufficient that, when
substituted in (258), (259), they give positive values for tan 2 c
and tan - C, or, which is the same thing, that a - b and A - B
will be of the same sign. This condition is both necessary and
sufficient. For a, b, A, being the given elements, denote by
B, c, C the other elements determined by the equations (257)(259). Now let us construct a triangle T, having the angle C
and the sides a, b, and calling A', B', c' the other elements of
this triangle, we have
tan C'= tan (a - b) s-n (a' + B') (258)
t 2 smn ( A ' (25) B)
tan 'tan B) sin 2- ('- B')'
tan i C= cot(A-B') sin - ( - b) ( )
sin a (a + b)'
tan (.A' + 2') tan (a + b)
tan (A' - B') tan - (a - b)'
and from (257) we infer
tan- ( A+ B) tan 2 (a + b)
tan (A - B) tan (a - b)'