348 Mr DAVIES on the Equations of Loci 



becomes, on account of (xxxi. 2.), 



d(p sec a (1.) 



and hence the length of the arc between the limits $' and <", is 



L = ((f)' cp") sec a (2.) 



agreeing in all respects with the commonly obtained result. 

 The area : this gives 



dA. = (1 cos0) dd = r , ' deb . tan a (3.) 



or integrating, 



A = 2 log sec 2 . tana (4.) 



When = 0, sec 2 = i, and log sec = 0. Hence, if we suppose 



the area to commence from the pole, and the radius-vector to revolve in the 

 positive direction, the whole surface traced out during the first quadrant of 

 of 0, will be = tan a log 2. 



XXXVI. 



Given the base and vertical angle of a spherical triangle, to find the 



locus of its vertex. 



Let a, /?, and a,, (3 a be the co-ordinates of the extremities of the base, 

 and 00 those of the vertex. Then, if we denote by d, 8 a 8 a the three sides 

 of any one of these triangles, and the given angle by e, we shall have (by II.) 

 for the determination of the locus, the four following equations : 



cos 8 cos a, cos a a + sin a, sin a,, cos (3, /? (1.) 



cos 8, =. cos <p cos a a + sin <j> sin a tl cos 6 /3,, (2.) 



cos 8 a = cos(f) cos a, + sin sin a, cos 6 /3, (3.) 



cos 8 = cos 8, cos 8 U + sin 8, sin 8 a cos 6 (4.) 



Inserting in (4.) the values of 8 8, 8 a from (1, 2, 3.), we shall have the 



2 



