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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.