Again, adding both members of (B) to unity, we have

But, b+c+a=2s, and b + c − a = 2 (sa).

159. To express the sines, cosines, and tangents of the halfsides of a spherical triangle in terms of the angles of the triangle.

Denoting A+B+C by 2 S, so that S is the half-sum of the angles of the triangle, we have B+C−A = 2 (S — A).

A).

(102)

sin b

=

sin C sin A

cos S cos (S- C)

(104)

sin A sin B

Again, adding both members of (C) to unity, we have

But A+B − C = 2(S – C), and A − B + C = 2 (S — B).

NAPIER'S ANALOGIES.

160. Dividing (99), Art. 158, by (100), we have

161. Multiplying (99) by (100), we have

162. Proceeding as in Art. 157, and applying (111) to the triangle A'B'C', we obtain,

But,

and

sin (A'+ B') tan c'
sin(4'- B') tan (a'-b')'

Whence,

=

† (A' — B') = † (180° — a — 180° + b) = {( − a + b),

† (a' — b') = † (180°— A—180°+B)={(−A+B).

=

tan (90° C)

cos (a+b) tan [180° - (A+B)]

Therefore, by Arts. 42, 46, and 47,

(113)

(114)