1884.] Non-Euclidian Plane Geometry. 91 



16. Consider a right-angled triangle ACB, where the points A, C 

 are on the meridian 0=0, and write (0 15 ; A 1 =cosec 2 ^ 1 ), (0 2 , 2 ; A 2 

 =cosec 2 2 ) (0 3 , 0; A 3 =cosec 2 3 ) for the points A, B, C, respectively. 

 Then if equations are 



for side BC, A 1 + B 1 (0 2 + A) + C 1 0=0, we have (^=0, 

 A 1 + B 1 (0 2 2 +A 2 )=0, A 1 + B 1 A 3 =0, whence 2 2 + A 2 =A 3 ; 

 for side CA, A2+B 2 (0 2 + A) + C 2 0=0, we have A 2 =0, B 2 =0; 

 for side AB, A 3 + B 3 (0 2 + A) + C 3 0=0, 



we have A 3 +B 3 (0 2 2 + A 2 )+C 3 2 =0, Ag + BgA^O. 



Observing that 1 =0 3 =0, we have 



cosh a = -J-T--== (0 2 2 + A 2 + A 3 ) , 

 * V A 2 A 3 



cosh &=T ' 



cosh c 2 y-^- (0 2 2 + A! + A 2 ) ; 

 or, reducing these by the relation 2 2 + A 2 =A 3 , they become 



\/A, \/A 3 A, , i _^A 3 Ao. 

 cosh a= =' whence sinh a= 7 "> tann a - ==r ^ > 



IT. 1 ' "8 



cosh 0=0 -7= = , 



A! + A S 



cosh c^ 



We have, moreover, 



v / C 1 2 -4A 1 B 1 ^C 3 2 -4A 3 B 3 ^/^A^j v / C 3 2 -4A 3 B 3 ' 

 which, writing A 3 = B 3 A 1? becomes 



cosB=- 



VA 8 v/C 8 2 -4A 3 B 3 ' 

 Or further reducing by means of 



