[528 
spec- 
50 
where Al, AJ are the (imaginary) tangents from A to the circle; or writing for 
shortness BI &c. instead of BAI, &c. (the angular point being always at A), 
consequently 
^sin BI. sin GJ 
sin G I. sin BJ 
where the numerator is 
sin BI sin (BJ — BG) — sin BJ sin (BI + BG) = sin BC sin IJ, 
or say = sin A sin IJ. Moreover taking the distance OA to be = sin^?, and the 
perpendicular distances from 0 on the lines AB, AC to be sine and sinb respectively, 
then if for a moment the angle IJ is put = 2a>, we have sin p sin co = 1: moreover 
sin BI sin BJ = sin (to — BO) sin (co 4- BO) = sin 2 co — sin 2 BO ; 
and sin p sin BO = sin c ; that is, sin BI sin BJ = 
sin Cl sin GJ = C0S , ^ ; also 
snrp 
sim c 
sm 2 p 
■ TT -a ~ • 2 £ cos p 
sin 1J = — sm ¿(0 = 2 sm co cos co = . ——- ; 
sin p smp 
whence the required formula 
j cos p sin A 
' ^ i* * 
cos b cos c 
In the same way, or analytically from this value, we have 
and thence also 
-r cos A + sm b sm c 
cos A = , 
cos b cos c 
■ cos p sm A 
tan A = 1 . ■ -,— . 
cos A + sm b sm c 
In particular, taking the line AG to pass through 0, or writing in the formula b = 0, 
we have tan BO = cosp tan BO = cosji tan 6; that is, BO = tan -1 . cosp tan 6; and similarly 
GO = tan -1 cos ji tan 6'; we ought to have A = BO A GO, that is, 
A = tan -1 cosp tan 6 + tan -1 cos p> tan 6' 
which, observing that sin p sin 6 = sin c and sin p sin 6' = sin b, also A = 6 + 6', is in fact 
equivalent to the above formula for tan A. 
Observe in particular that when A is at the centre, p is =0, and the formula 
becomes A = 0+0', =A, or say for an angle at the centre, 0 = 0.