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

62 Solution of Spherical Triangles.
To find C we have, from (159),
Itan - c = {-tan -(a+ b)} + [-cos(A+ B)} -1 cos(A-B).
Now
- tan 2 (a + b)} = 8603353, {- cos (2 + B)} = 1-5475498,
I cos - (A - B) = 1i9978668.
Hence tan -c =-4100083;.. c= 137~ 29' 3".
Observation.-In the foregoing calculation it is seen that,
when an angle is between 90~ and 180~, we have the sign
minus before its cosine and its tangent, the reason of which
is obvious.
Or thus:-Let fall the perpendicular BE; then denoting the
arc AE by 0, CE will be b - 0. Then, from the right-angled
triangles, we have
tan (b - 0) = tan a. cos C, tan A/tan C = sin (b - 0)/sin 0.
B -4
Fig. 20.
The first equation determines 0, and the second A. In a similar
manner B may be found. Lastly, from the same triangles, we
have cos c/cos a = cos 0/cos (b - 0). Hence c is found.
EXERCISES.-XIX.
1. Given a= 88~ 12'20", b = 124~ 7'17", C =50 2' 1";
find A, B, c.
2.,, a=110 55 35, b= 88 12 20, C=47 42 1;
find A, B, c.
3., a= 65 15 12, b= 47 42 1, C= 59 4 25;
find, B, c.