420 
123. To express 0 , <£, and \p in terms of the polar distances 
and longitudes. 
Then, according to § 110, if we substitute cos p x for 1, 
cos p 2 for m, and cos p 3 for n, 
We have for the faces nm 1 and mn 1, or 
cos p 3 , cos p 2 , cos p v and cos p 2) cos p 3 , cos p v 
and cos 6 = cos p 2 cos p 3 + cos p 2 cos p 3 + cos 3 p x 
= 2 cos p 2 cos^g+cos 3 p r 
For the faces m 1 n , and 1 m n , or 
cos p 2 , cos p v cos p 3i and cos cos p 2 , cos j:> 3 , 
and cos <p = cos p x cos p 2 +.cos yq cos p 2 + cos 3 p 3 
= 2 cos yq cos p 2 + cos 3 p 3 . 
Also for the faces- m 1 n and m 1 n , or 
cos p 2i cos jq, cos p 3 , and cos p 2 , cos p v cos £> 3 , 
cos \p = cos 3 p 2 + cos 3 p x — COS 3 jq. 
But referring to § 118 cos 3 p 2 = sin 3 _p 3 sin 3 A 3 
and cos 2 p x = sin 3 p 3 cos 2 A 2 . 
Hence cos ip = sin 3 p 3 sin 3 A 3 + sin 3 p 3 cos 3 A 3 — cos 3 p 3 
= sin 2 p 3 — cos 3 p 3 = 2 sin 3 2 ^ 3 — 1 • 
And 1 + cos \p = 2 sin 3 p 3 . 
Therefore 2 cos 3 ^ = 2 sin 3 p 3 , 
u 
and 
Whence ^ — 
cos 
| = sin ^ 3 = cos (90° -ft). 
90° — p 3 , or \fs = 180°— 2p 3 , 
This result might have been obtained at once by inspection 
from (fig. 31*, Plate IV.*) Forp 3 is the north polar distance of 
the face 1 mn, and 180° — p 3 that of 1 mn. The poles of both 
these faces also lie in the same meridian. 
Hence <p = 180° -p 3 — p 3 =180°- 2 p 3 . 
Again, using the formulas § 119, 0 is the inclination of the 
pole of the face mn 1 to that of nm 1, p x the north polar 
distance of the pole of mn 1, and \ its longitude referred to 
G 1 as north pole, and C 3 C 2 C 5 as equator and measured from C 3 . 
p x the north polar distance of nm 1 and 90 — A x its longitude 
referred to the same north pole and equator. 
Hence cos 6 = cos p x cos^q + sin^q sin^q cos (90-2 AJ 
= cos 2 jq + sin 3 p x cos (90 — 2A : ) 
= 1 — sin 3 jq-b sin 3 p x cos (90— 2 A : ) ; 
and 1 — cos 6 = sin 3 2 % {1— cos (90 — 2 A x ) }. 
f) 
Therefore 2 sin 3 -=2 sin 3 p x sin 
e 
90-2 A x 
and sin |-~sin p x sin (45 — AJ. 
2 
