90 
Therefore 
kBsin ABsin BAP= ly sin CA sin CAP. 
Tn like manner 
ly sin BCsin CBP =hasin ABsin ABP, 
and hasin CA sin ACP =k8 sin BC sin BOP. 
Hence, 8sin ABsin BAG =ysin CA sin CAG, 
ysin BC sin CBG =asin AB sin ABG, 
asin CAsin ACG =Bsin BOsin BCG. 
Therefore 
sin BAP _,sin CAP 
sin BAG sin CAG’ 
sin CBP =F sin ABP 
sin CBG sin ABG’ 
sin 4 CP ail sin BCP 
sin ACG sin BOG ° 
The point P will be called a pole, and will be denoted by 
the symbol of the ray intersected in that point by the surface of 
the sphere. The points A, B, C, @ are the poles 100, 010, 
001, 111 respectively. 
l 
Zone-circles. 
21. In fig. 8 let A, B,C, GP be the poles 100, 010, 
001, 111, wow respectively. Let a great circle passing through 
P meet the great circles BC, CA, AB in D, E, F respectively. 
Then (Townsend’s Modern Geometry, 82, Cor. 3), having regard 
to the signs of the six arcs, 
sin DP sin FE + sin PF.sin DE + sin FD. sin PE=0. 
Therefore 
sin FE sin DP sin DE sin oe 10 
sn /D sin PH © sin FD sin PE ae 
whence 
sin ACF sin BCP sin DAC sin BAP 
sin FOB sin AOP* sin BAD sin CAP 
+1=0. 
