luong@tonto.DEC (Van Luong Nguyen UHO DTN 264-6560) (07/17/84)
>Given right triangle ABC (A being the right angle), with point D on AC such >that BD=CD=1, and AF being the perpendicular from A to BC. If CF=1, find AC. Solution: Let angle ACB = x -------- Since angle AFC is 90 degrees, we have : CF/AC = sin x therefore, as CF=1: AC = 1/cos x (1) Now, DC=DB, therefore angles DBC=DCB=x, and it follows that: angle BDA = 180 - BDC = 180 - (180-2x) BDA = 2x Therefore: AD/BD = cos 2x Since BD=1 and cos 2x = [2(cosx)(cosx) - 1] , AD = [2(cosx)(cosx) - 1] and: AC = AD + DC = [2(cosx)(cosx) - 1] + 1 = 2(cosx)(cosx) (2) Combining (1) and (2) gives: 1/cosx = 2(cosx)(cosx) or: cosx = (1/2)**(1/3) and finally: AC = 1/cosx = 2**(1/3) (third root of 2) = 1.25992.... Van Luong Nguyen, Digital Equipment Corporation, Nashua, NH.