417 
but spherical triangle fG s g is measured by arc kC 2) and fgC 3 
is 90°; hence 
= and S inA 1= ^LP3 
sm jp x sin 90° sin p x 
From the spherical triangle G x G 2 f } we have 
cos GJ= cos G X G 2 cos GJ+ sin G X G 2 sin GJ cos G 2 G x f } 
or, cos ^ 2 = cos 90° cos^ 3 + sin 90° sin_p 3 cos (90° — X 3 ) 
= sin jp 3 sin X 3 . 
From the spherical triangle G 2 fg, we have 
sin GJ _ sin G 2 gf or sin jp 2 _ sin 90° 
sin fg sin fG 2 g sin 90° -p 3 sin X 2 
and sin X 2 =- 
sm p 2 
Hence cos p x = sin cos X 3 , sin X x = -° s ; 
sin 
and cos _p 2 =siny> 3 sin X 3 , sin X 9 = cos jp 3 
sin jp 2 
119. To find the angle between the poles of two faces in 
terms of their polar distances and longitudes. 
Let G X F be the polar distance of F (fig. 41, Plate IV .*), 
G 3 L its longitude, G x f the polar distance, and Col the longitude 
of /. 
Also let C' 1 P=P 3) GJ = }h ; 0,1=1,, C,l=\,, and F/=0. 
1 hen m spherical triangle G x Ff 
cos Ff— cos G X F cos GJ+ sin G X F sin GJ sin FGJ. 
Then angle FGJ is measured by arc Ll= LG 3 —IG 3 . 
Hence cos 0 = cos P 3 cos^ 3 + sin P 3 sin^ 3 cos (L 3 — X 3 ). 
To adapt this to logarithmic computation — 
cos 0 = cos jp 3 {cos P 3 + sin P 3 tan p 3 cos (P 3 — X 3 )}. 
Let tan a = tan £> 3 cos (P 3 — X 3 ). 
Then cos 0 = cos_p 3 {cos P 3 + tan a . sin P 3 } 
cos p 3 
cos a 
COS J ) 3 
cos a 
{cos P 3 cos a-f sin P 3 sin a} 
cos (P 3 a). 
120. To find the distance between any two poles on the 
sphere of projection in terms of the three polar distances 
from G v G 2 , and G 3 . 
§119. cos 0 = cos P 3 cos jg 3 -j- sin P 3 sin^ 3 cos (P 3 — X 3 ) 
= cos P 3 cos j:> 3 + sin P 3 sin_p 3 (cos l 3 cos X 3 
+ sin L 3 sin X 3 ) 
= cos P 3 cos £> 3 + sin P 3 sin _p 3 cos L 3 cos X 3 
+ sin P 3 sin jp 3 sin L 3 sin X 3 ; 
but § 118, cos 4^= sin cos X 3 cos P x = sin P 3 cos L 3 
cos _p 2 = sin_p 3 sin X 3 cos P 2 =sin P 3 sin L y 
