422 
and sin X. 
cos 2. 
sec 45° sin ~-\ sin 
2 2 ) 
sec 45° sin 
0 
1 + 
• $ 
sin 2- 
2 
cos 
2 
sec 45 sin 
sin i cos 45 
Let tan 3 a— ^ 
Then sin X Q = 
sin 
c 
. 9 
sm - 
2 
9 
{1 -f tan 3 a} — - 
. 9 
sm - 
2 
cos 45 cos cos 3 a 
2 
cos 45 cos ^ 
2 
126. Given 0 and 9 , find p 3 and X 3 . 
(Fig. 42*, Plate IV*.) — Let be the pole of 1 mn, a 2 
that of 1 n m } and a 3 that of m 1 n. 
Join a v a 2 by arc of great circle cutting oG 3 in f f 
and a v a, 3 by arc of great circle cutting od in e ; 
also G v a 1 by C 1 a x cutting dG 3 in g, 
and 0a 1 cutting dC 3 in h. 
Then G^ ==p 3 , C 3 g—\ 3 , 0 1 o = 54° 44', and G 2 od=6 0° ; and 
let oa^ — P, G 2 oa 1 —L. 
j Q 
Also a 3 a x = <j> an d ea x — - a x a 2 = 9 ct x f= - 
iL — 
From spherical triangle oa^f 
sm 
therefore 
9 
2_ sin P 
sin L sin 90 c 
sin a Y f _ sin oa x 
sin a x of sin ofa x 
Also in spherical triangle oa x e 
sm a,e 
sm oa-. 
sm eoa-, 
sm oea-t 
<P 
and 
sm - 
Hence ^ — 
sm L- . -r, 
2 __ sm P 
sin (60° — L) sin 90 c 
t 
2 
• 9 
sm Z 
and 
sinL sin (60 —L) 
. 0 
2 _ sm L 
sin L 
• <p sin (60° — L ) sin 60° cos L — cos 60° sin L 
Sm 2 
