31151029036484.pdf

248 SPHERICAL TRIGONOMETRY.
171. Again, from (310) and (312) we find
— tan iJ 4- tan R' + tan R" + tan R'"
_ cos S + cos (S — A) + cos (S — B)+ cos (S •
= ~ N
_ 2 cos ^ A cos i (B + C) + 2 cos ^ A cos i (B
~ N
whence
4 cos J A cos J B cos } G
— tan iJ 4- tan R' + tan R" + tan R'" =
(315)
We shall find in a similar manner
4 cos } ^ sin } 5 sin J C
tan R — tan R' + tan R" + tan R'"
If
4 sin J J4 COS ^ B sin^ G
tan B + tan i2' — tan R" + tan R'" =
(316i
If
4 sin J ^ sin J B cos J 0
tan .K 4- tan R' + tan i2" — tan R'" =
N
It is also easily shown that
tan' R + tan' R' + tan' R" + tan' i2"' = 2 4- 2 cos J1 COS B COS (7 (317)
172. To find the radius of the circle inscribed in a given spherical triangle.
Kg. 28. ^ In Pig, 28, 0 being the pole of the required circle,
draw OP', OP" and OP'" to the points of contact, and join
OA, OB. We have OP" = OP'" and the triangles
ip, A OP" and A OP'" right-angled at P" and P'"; hence
Bin OAP" =
sin OP" sin OP'"
sin ^ 0 sin .4 0 = sin0.4P"'
therefore 0 A P" = 0 A P'", (for we cannot have
0A F" := TT — OA P'"), and the pole of the inscribed circle is consequently found
by the same construction as inplano, namely, by bisecting the angles of the triangle.
If then we put s z=^ ^ (a + b + c), and r = radius of the inscribed circle, we
have
AP'" + BP' + GP' = AP'"+ a = a, AP'" = s — a
and the right triangle A 0 P'" gives
tan r = sin (a — ffl) tan } A (818)
corresponding with the formula of PI. Trig. (288).
Substituting, in (318), the value of tan } A,
^sin (s — a) sin (s — 6) sin (s — c)\
sin s /
tan r -.
'J(c
tan r = •
Substituting, in (318), the value of sin (s — a) given by (58),
^^^ ^
^
2 cos J A cos J B cos J G
Also, by (51), we have iV = } sin B sin G sin a, which reduces (320) to
Bin i B sin i C .
tan r = =—=~-r^— sm a
'.OS ^ A
(319)
(320)
(321)