95 
cot AP =- 7 00S BAP= tan E Ecos CAP, 
cot BP = 2 i cos CBP = ; cos ABP, 
cot CP =— ‘cot J cos ACP = i Leg Ecos BCP, 
i sa 
(tan CP)? = (tan £)’, 
The arc H may be taken for the element of the crystal. 
Position of any pole in the rhombohedral system. 
30. In this system the arcs joining every two of the poles 
100, 010, 001 are all equal, and the arcs joining the pole 
111 and each of the poles 100, 010, 001 are all equal. 
In fig. 10, let A, B, C, O be the poles 100, 010,001,111 
respectively; P the pole hkl. Let the zone-circles OA, OB, OC 
meet the zone-circles BC, CA, AB in D, E, F. Then, since 
BC=CA=AB and OA = OB= OC, it is evident that BC, CA, 
AB are bisected in the points D, HL, F; that OD=OEH=OF; 
that the angles at D, H, F are right angles; that the symbols 
of D, H, F are 011, 101, 110 respectively ; and that the six 
angles having their apices in O are each of 60°. 
The symbol of OA is 011, and that of OB is 101. 
Therefore (27), 
cot AOP — cot AOB_h-l 
cot AOL —cot AOB k—-l° 
But AOB=120’, AOF=60°. Therefore tan AOB=— 1/3, 
tan AOF= 1/3. Hence 
ees VT) V3 
tan AOP=—7 5 . 
