The Rev. H. Moseley on Conchyliometry. 303 



angle, since the plane I P H is a tangent to the cone, and 

 IPL passes through its axis, 



.•. cos b e = cos a e . cos a b. 



Now£<> = LPH = «,fle=LPI = RIP=, 

 fl 6 = IPH = A, 

 .*. cos a s= cos i . cos A (3.) 



.*. i + tan 2 « = sec 2 a = 5 s-r 



qos z i cos a A 



2 1— cos 2 » cos 2 A _ sin 2 A + sin 2 < cos 2 A 



cos 2 1 cos 2 A cos 2 < cos 2 A 



tan 2 A 



cos* * 



a tan 2 1 o 1 



+ tan 2 i= - A . o +tan 2 t=^ 3 . +1 Uan\ 

 cot 2 A sin 2 » [ cor A sin 2 1 ^ 



Now cot A sin 1 = — Jp — (equation 1.), 



" = { 1+ (isir§)T tan (*J 



2* 



r. / 2 



.•. tan 



Similarly, it may be shown that 

 sin 



in--{l + (^.)'}*rinA (5.) 



III. The Area of a Conclwidal Surface. 



Let R P S represent any position of the generating curve, 

 and QPm a portion of one of the spiral lines generated by 

 any point P in it. Let I represent the apex of the cone on 

 whose surface the spiral P Q is described. Join P I and 



draw P H a tangent to the spiral, and P T a tangent to the 

 generating curve in P. Imagine a sphere described with 

 radius unity from the centre P, and let a b, be, ac repre- 

 sent the intersections of the planes H P I, H P T, and I P T 

 with its surface. Now the plane H P I, being a tangent to 

 the cone at P, is perpendicular to the plane RIP which passes 

 through its axis I R S ; the spherical angle b a c is therefore a 

 right angle. 



Moreover, the angle I P Q made by a tangent to the spiral 

 with the line I P drawn from the summit of the cone is, in the 



