412 
ON THE N ON-EUCLIDIAN GEOMETRY. 
[528 
I return to the expression for cosh a; in explanation of its meaning, let the 
distances OB, 00 be q, r respectively and let the angle BOO be a; to find q we 
have only to take 0 at 0, that is, in the formula for cosh a to write r = 0, we thus 
1 1 
find cosh q = : and similarly cosh r = , whence also 
* cos q J cos r 
cos q = sech q, sin q = i tanh q, 
cos r = sech r, sin r = i tanh r, 
also, as seen above, a = a; the formula thus is 
. _ 1 + tanh q tanh f cos a 
cosh a = t-== r—= 
sech q sech r 
= cosh q cosh r + sinh q sinh r cos a, 
or, what is the same thing, it is 
cosh a, — cosh q cosh r 
cos a = . . _ . t _ 
smh q sinh r 
viz. as will presently appear, this is the formula for cos BOG in the triangle BOG. 
From the above formulae 
and 
. _ 1 — sm q sm r cos a. 
cosh a = 
cos qcos r 
. -r cos v sm A —r cos A 4- sm b sm c 
sm A = —±7 , cos A = t 
cos b cos c cos b cos c 
and the like formulae for b, c, B, G, it may be shown that in the triangle ABG we have 
. _ cos A + cos B cos G 
cosh a = ; —=— ; —= . 
sin B sin C 
In fact, substituting the foregoing values, this equation becomes 
(1 — sin 2 a) (cos A + sin b sin c) + (cos B + sin c sin a) (cos G + sin a sin b) _ 1 — sin q sin r cos a 
sin B sin G cos b cos c cos q cos r 
that is, 
cos A + cos B cos G — sin 2 a cos A + sin a sin b cos B + sin a sin c cos C + sin b sin c 
= sin B sin (7(1 — sin q sin r cos a), 
or, what is the same thing, 
sin 2 a (cos B cos G — sin B sin G) + sin a sin b cos B + sin a sin c cos G + sin b sin c 
= — sin B sin G sin q sin r cos a, 
that is, 
(sin a cos B -l- sin c) (sin a cos G + sin b) = sin B sin C (sin 2 a — sin q sin r cos a), 
a relation which I proceed to verify.