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

First Class. 25
sin B 2n
In like manner, =: -
sin b sin a sin b sin c
sin.A sinB sin C
Hence. =.-, (47)
sin a sin b sin c
and the proposition is proved.
Or thus:-Let ABC be the triangle, O the centre of the
sphere. From any point D in OA draw DG perpendicular to
the plane BOC; and from G draw GE, GF at right angles to
OB, OC. Join DE, DF. Now since DG is perpendicular to
the plane, and GE perpendicular to OB, a line through G
parallel to OB would be perpendicular both to DG and GE,
A
B
Fig. 13.
and therefore normal to the plane DGE. Hence the angle
DEG is equal to the spherical angle B (~ 8). In like manner,
DFG is equal to C. Now DE sin DEG = DG = DF sin DFG;
therefore DE sin B = DF sin C; but DE = OD) sin D OE = OD
sin c, and 1DF= 02 sin DOF= OD sin b. Hence
sin B sin c = sin C sin b,
or
sin B: sin C:: sin b: sin c.