In like manner, since p 3 and X 3 ; and_p 3 and 90 — X 3 , are the 
polar distances and longitudes of the faces 1 m n and m 1 n 
referred to G x as north pole, and G 3 G 2 G 5 as equator, 
cos (J) = cos jj> 3 cos p 3 + sin jr? 3 sin p 3 cos (90 — 2 A 3 ), 
which gives as above 
^ — sin p 3 sin (45 — X 3 ). 
sm 
1 24. Given $ and i//, find p 3 and X 3 . 
We have seen, § 123, thatp 
80 -| ; 
also sin ^ = sin ^> 3 sin (45 — X 3 ), 
u 
sin t 
therefore sin (45 — X 3 ) = — 
sm p 3 
125. Given \p and 0, find_p 3 and X 3 . 
§123. Pa =90-t. 
Q 
sin ^=sin (45 — \) sin_p 1( 
but sin 45 = cos 45 = 
0 
= (sin 45 cos \ — cos 45 sin X x ) sinp 1 ; 
1 
\/2 
y/% sin -=sin p x cos A x — sin p x sin A r 
U 
Referring to (fig. 40*, Plate IV.*), and remembering 
§117, that p x =fG 3 p 2 =fG 2 p 3 =fC 1 
\ = G 2 h \ 2 =0 3 h \ 3 =G 3 g. 
From the spherical triangle fcj G v we have 
si - sin/r/ or sin p l _ sin (90- p 3 )_ cos p 3 
sin Xj sin \ 
from 
sin fg G s sin f G 3 g sin 90 
Therefore sin^q sin X 1 =cosp 3 . 
Also from spherical triangle G x fG 3 , we have 
sin f G 3 _ sin f G 1 , sin p 1 __ sin p 3 
sm p 3 
cos \ Y 
sin fG 1 G 3 sin fG 3 G L sin X 3 sin (90 — A-J 
Therefore sin p x cos \ = sin p 3 sin X 3 . 
Hence 
9 
\/ 2 sin ~=sinp 1 cos X x — sin^q sin X 1 =sinp 3 sin X 3 — cos p 3 
U 
— sin ^90—^ sin X 3 — cos ^90— ^ = cos ^ sin X 3 — sin ^ 
Hence cos ^ sin X 3 = >/2 sin ? -f sin ^ = sec 45° sin - + sin 
