288 Mr DAVIES on the Equations of Loci 



Draw LR perpendicular to LD, and make it = ; and join PR. Then 



R is the centre of the circle LD, and we are required to find the co-ordi- 

 nates of that point. 



7T 



Now PLR = g e, and PLR is a quandrantal triangle. Hence 



cos X = sin a sin e (3j 



and cosPLB, or cos(/c ft) = cot a cot X (4) 



or K = ft cos- 1 ( cot a cotX) (5) 



From (3) we obtain, cot X = cot cos" 1 (sin a sin e) 

 sin a sin e 



sin 2 a sin 2 e 



and.-.tanX = 



sin a sm e 



(6) 



It will hence follow, that 



cos /c = cos jS cot a cot X + sin ft \/l cot* a cot* X 



cos ft cos a sin e sin ft cos e 



Vl sin 2 a sin 2 e l sin 2 a sin 2 e 



_ sin ft cos e sin e cos ft cos a 



sin 8 a sin 2 e 





/, -- 5 cos e cos /3 + sin 6 sin 6 cos a 

 sin/c = vl cos 2 /c = - ^ ^ - ...... (8) 



v 1 sin 2 a sin 2 e 



For sin/c, cos/c, tanX, in equation (1), write the values furnished by 

 (6, 7, 8) ; then, 



i cos#(sin/3 cose sine cos/3 cosa)- 

 cot o> = cosec a cosec e ? /j , /, . . V I ....... (9) 



\ + sm0(cos0 cose + sine sm/3cosa)J 



