31151029036484.pdf

SOLUTION OF SPHE.IICAL OBLIQUE TRIANGLES.
95. CASE V. Given the three sides, or a, b
and e. (Fig. 9.) We have three methods
for computing the half angles :
ist. By the sines, from (31), remembering
that
s = 1 (a 4- 6 4- e)
sin IA=\ ('^^^i^3^{LriA\
^ \ sin 6 sm e /
sin J 5 •
sin i C
2d. By the cosines, from (33),
sin (s — c) sin (s — a)^
sin e sin a /
sin (s - a) sin (s — 5)^
sin a sin b
Mg. 9.
203
(164)
cosi^= //Z^^^^^MfJ^N
^ >/ \ sin 6 sin e /
sin (s — 5)
cos J 5 >J V SI
sm e sm a
cos J'^-JC
sin s sin (s — e)
sin a sin 6
3d. By the tangents, from (34),
tan IA _
/ , sin (s — b) sin {s — c)\
*>* V sin s sin {s-a) '
tan»5=/(^-^f-^)f^(\-^))
b)
tan|C=/C(f~^)^;°(^7Jl)
^ V sin sm s .s sin am (« ix — c) el
'
(165)
(166)
When only one of the angles is required, the simplest method will
be by (165), but if the required angle is less than 90°, it will be
found more accurately by (164), for then ^ A