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

64 Solution of Spherical Triangles.
Draw CD perpendicular to the plane A OB, also DA', DB' perpendicular to OA, OB. Then it is evident that CA', CB' are
also perpendicular to OA, OB.
Let B'OD = a, DOA'= 3, DOC= 8;
then we have
CB'=sina, OB'= cos a; CA'=sinb, OA'=cos b;
CD = sin 8, OD = cos 8; CAD=' - A,*CB'D = B.
The triangles ' OD, B'OD, A' CD, B' CD, have the angles
A' B' D right;
OB' OA'
DO = cos - =
co Bs3OD cosA' OD'
cos a cos b
Hence s (272)
cos o coQ B'
DC = sin 8 = CB' sin C7'D>= CA'.sin CA'D.
Hence sin a. sin B = sin b sin A. (273)
DB' = OB' tan B'OD = CB' cos CB'D.
Hence tan a = tan a cos B. (274)
D.' = OA' tanA'OD = CA' cos CA'D.
Hence tan / = tan b cos 4. (275)
From (272) we get
tan 2 (a + b) tan ( (a - b) = tan - (a + /) tan - (a -,);
but (a + /3) = c. (276)
Hence
tan 1 (a + b) tan i (a - b) = tan i G. tan } (a - f). (277)
The formulae (272)-(277) solve all the cases of obliqueangled triangles.
Ist Case.-Given the three sides.
(276) gives (a + /), (277) (a - 3), (274), (275) give A, B.
2nd Case.-Given the sides a, b, and the angle A.
(273) givesB, (274), (275) give a, /, (276) gives c, and (273) C.
3rd Case.-Given the angle A and the adjacent sides.
(275) gives /, (276) a, (272) a, and (273) determinesB and C.