31151029036484.pdf

£34 SPHERICAL TRIGONOMETRY,
the difference and sum of which give
sin a cos (S 4- J AB) sin J AB = sin A cos (5 -4- i AJ) sin J Ai
sin a sm (-B 4- J AB) cos J Ai? = sin A sin (S 4- J A5) cos J A5
from which, by division, we find
tan J A5 _ tan (S 4- J A5)
tan i AB ~ tan (54-1 AB)
(2391
and in the same manner
tan J Ac _ tan (c 4- ^ Ae)
tan J AC ~ tan (C4- i AC)
(240)
The product of (237) and (239) gives
whence also*
sin I Ab ^ _
tan J AC
sin {b -\- ^ Ab) cos | Ac
cos (e 4- J Ac) tan (-B 4- J AB)
(241)
sin J Ac
tan ^AB
_ sin (c -f I Ac) COS | .A5
cos(6 4-J A6)tan(C4-JAC)
(242)
Fig. 24.
a 136. CASE III. b and c constant. The given
c-,triangle being ABG, Fig. 24, the parts of the
derived triangle ABC are J, c, a -f Aa, S -f AB,
C4- AC, A 4- AJ.. Joining C G' we have in B C C,
ty (42),
sin (a 4- J Aa): sin J Aa = cot | AB : tan ^{BCG'—B G'G)
But observing that J. C = J. C, J. CC = J. C'C, we have
BCG' = AGG' - G
BC'C = AC'C + O-if AC
i {B OC -BG'0) = -{Q+l AG)
and the above proportion gives, therefore.
sin I Aa _
tan J AB
sin (a -f J Aa)
cot (C 4-i AC)
(243)
* The equations (239), (240), (241), and (242), contain each two factors less thin
the corresponding equations given by Cagnoli.