VOL. LXII.] PHILOSOPHICAL TRANSACTIONS. 349 



EFc ; and 2afi Is greater than af^ — ai^ ; also af' — ai'^ together with efc, is 

 equal to eic ; therefore by composition, the ratio of ib to bf is greater than that 

 of Eic to EFC; and the ratio of ac X ib to ac X fb greater than that of eic to 

 efc: also, by permutation, the ratio of ac X ib to eic greater than the ratio of 

 AC X fb to efc. But the first of these ratios is the same with that of the square 

 of the radius to the rectangle under the sine of bah and the tangent of abh ; and 

 the latter is the same with that of the square of the radius to the rectangle under 

 the sine of bag and the tangent of abg; therefore the latter of these two rect- 

 angles is greater than the other. 



Again, let the triangle bak be formed, with the angle bak less than bag, and 

 the perjjendicular kl be drawn. Then the rectangle under the sine of bak and 

 the tangent of abk, is to the square of the radius, as the square of kl to the 

 rectangle under ac, bl. Here, pl being to fb as 2afl to 2afb or efc, and 

 2afl less than al^ — af", by conversion, the ratio of lb to fb will be greater 

 than the ratio of elc to efc ; therefore, as before, the rectangle under the sine 

 of bag and the tangent of abg is greater than that under the sine of bak and the 

 tangent of abk. 



Corol. I. bp is equal to the tangent of the circle from the point b; therefore 

 BP is the tangent, and ab the secant, to the radius ac, of the angle, whose co- 

 sine is to the radius as ac to ab. Therefore af is the tangent, to the same ra- 

 dius, of half the complement of that angle; and af is also the cosine of the 

 angle bag to this radius. 



Corol. 2. The sine of the angle composed of the complement of agb, and 

 twice the complement of abg, is equal to 3 times the sine of the complement of 

 AGB. Let fall the perpendicular ah, (fig. 5), cutting the circle in i ; continue 

 gf to K, and draw ak. Then bf' = ebc = gbl. Therefore gb : bf :: bf : bl, 

 and the triangles gbf, fbl are similar. Consequently fl is perpendicular to gb, 

 and parallel to ah; whence gh being equal to hl, gm is equal to mf, and mk 

 equal to 3 times gm. 



Now the arc ik = 2ic + gi; and the angle iak = 2iac -f gai; also gm is 

 to MK as the sine of the arc gi to the sine of the arc ik, that is, as the sine 

 of the angle gai to the sine of the angle iak. Therefore the sine of the angle 

 iak (= 2iAC + gai) is equal to 3 times the sine of the angle gai; but gai is 

 the complement of agb, and iac the complement of abg. 



Corol. 3. If (fig. 6) any line bn be drawn to divide the angle abg, and an be 

 joined, also ao be drawn perpendicular to bn, and continued to the circle in p, 

 the sine of the angle composed of nap and 2pac will be less than 3 times the 

 sine of the angle nap. Draw nqr perpendicular to ab, cutting ap in s ; join 

 AR, and draw qt perpendicular to bn, and parallel to ao; then sa'^ = nbt. But 

 Ba' is greater than the rectangle ebc, that is, greater than the rectangle nbv. 



