[ 43 6 ] 
AC, BL. Here, FL being to FB as 2 AFL to 
2AFB or EFC, and 2 AFL Iefsthan AL? — AF?, 
by converfion, the ratio of LB to FB will be greater 
than the ratio of ELC to EFC$ therefore, as be- 
fore, the re&angle under the fine of BAG and the 
tangent of ABG is greater than that under the fine, 
of BAK and the tangent of ABK. 
Corollary i. 
BF is equal to the tangent of the circle from the 
point Rj therefore, BF is the tangent, and AB the 
iecant, to the radius AC, of the angle, whofe cofine 
is to the radius as AC to AB. Therefore, AF is 
the tangent, to the fame radius, of half the comple- 
ment of that angle ; and A F is alfo the cofine of 
the angle BAG to this radius. 
C o r o l. 2. 
The fine of the angle compofed of the comple- 
ment of AGB, and twice the complement of AB G» 
is equal to three times the fine of the complement 
of AGB. Let fall the perpendicular AH (Fig. 2.), 
cutting the circle in I ; continue G F to K, and draw 
AK. Then BF? = EBC = GBL. Therefore, 
GB : BF :: BF : BL, and the triangles GBF, 
F BL are fimilar. Confequently F L is perpendi- 
cular to GB, and parallel to AHj whence GH 
being equal to HL, G M is equal to M F, and 
M K equal to three times G M. 
Now, the arc I K = 2 IC -4- G I ; and the angle 
I A K ~ 2 I A C 4- G A I ; alfo G M is to M K as 
the 
