of Edinburgh, Session 1879 - 80 . 
663 
Whence do tan B = - p tan — c£B .... (11) 
P 
From these equations we get successively: 
• T) • C » & 
sm B sm— = sm — 
P P 
I. 
a b c 
cos — cos — = cos — ... 
P P P 
. II. 
sin A cos — = cos B ... 
P 
. III. 
tan — cot — = cos A ... 
P P 
. IV. 
For hyperbolic space we get in like manner 
sin B sin h — = sin h — 
P P 
r. 
, a , b 7 c 
cos h — cos h — = cos h — 
P P P 
. ir. 
sin A cos h — = cos B ... 
P 
. nr. 
tan h — cot h — = cos A ... 
P P 
. IV'. 
The reader will observe that these are simply the formulae included in Napier’s 
rules for right-angled spherical triangles. The only modification being that in 
hyperbolic space hyperbolic functions take the place of circular functions. 
In other words, the trigonometry of single elliptic space is identical with 
the geodetic trigonometry of a sphere, although it would not be correct to say 
that the planimetry of single elliptic space is identical with the geodesy of a 
sphere. 
For hyperbolic space the analogue is the pseudo-spherical surface of Bel- 
trami. 
Parallels. 
As an illustration of the application of the above formulae to parallels, I 
shall find the parallel angle in hyperbolic space. 
Taking formula IV', if we make B move off to an infinite distance, then AB 
becomes the parallel to CB. A is then the parallel angle corresponding to b. 
Now since c = oo we have 
cot h — — 1 , 
therefore 
U A 
0 0 
A 4. 7 b eP - eP 
cos A — tan h — = b b 
(12) 
eP + eP 
VOL. X. 
