SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 196

To find a. We have several methods : 1st, directly by (3V
which gives

sin 6 sin ^
sm a = — : — ^— (158)

sin ^ ^ '



2d« By (146), where $ and ^' have the same values as in this case,

S $' COJ
COS $



COS $' COS h



8d. By (150),

cot a = cot <p' cos 5 (^60)

4th. By (147), where 9- and 9-' have the same values as in this
case,

cosy cot 6

cot a =» K — (161)

cos 9- ^ '

5th. By (152),

cos a = cot y cot B (162)

Each of the last four formulae gives two supplemental values of
a by the two values of ^' or 9-', employed in the second members.
From (156) we have

cos A = tan ^ cot h

which with (159) gives

cos a sin I

-—-J =5 cos $' X -^—1
cos A ^ sm ^

The sign of the second member of this equation depends upon
that of cos ^', since sin h and sin ^ are always positive. Hence
when cos ^' is positive, cos a and cos A must have like signs ; and
when cos $' is negative, cos a and cos A must have different signs.
