97 
The arcs AG, BG, CG meet the arcs BC, CA, AB in the 
poles 011,101,110. Putting 010,011=D,001,101=E, 
100,110=F, and observing that the angles A, B, C are right 
angles, the equations in (20) become 
tan D, tan CBP = be #, tan ACP= E tan F, 
i 
tan BA P=— i z 
k 
cot AP= : cot F cos BA P= f tan H'cos CAP, 
cot BP= F cot D cos CBP= : tan F’'cos ABP, 
cot CP= 7 cot Hos AOP =, tan D cos BCP. 
v 
Any two of the ares D, H, F'may be taken for the elements 
of the crystal. They are connected by the equation 
tan D.tan LH. tan f= 1. 
Position of any pole in the oblique system. 
32. In this system the arc joining the poles 100,010, 
and the arc joining the poles 010, 001 are quadrants. Let 
A, B, CO, G, P be the poles100,010,001,111,Ak1 ve- 
spectively. Let BG, BP meet CA in L, S&. Then ZL will be 
the pole 101, and S the pole hOJl. The arcs AB, BC are 
quadrants, and consequently the angles ACB, CAB are right 
angles. Hence the equations in (20) become 
famCAe ie sm Alsi CS i) tan ACP 
tan CAG 1’ sm OLsin AS 1’ tan AOG i’ 
Putting 
Asin CL 
oN n nar 
the arc AS will be given by the equation 
tan (AS—4AC) =tan $AC tan (17 — 80). 
