In triangle ABC, angle C has a measure of 60 degrees. Point D lies on side AC so that BD bisects angle B and BD = 1. Similarly, point E lies on side BC so that AE bisects angle A and AE = 2.
Find the area of triangle ABC.
It is easy to verify that if B is a right angle,
then AE = 2BD.
Applying the law of sines to triangle BCD gives,
BC BC BD 1
 =  =  = 
sin(75) sin(BDC) sin(BCD) sin(60)
or
sin(75)
BC = 
sin(60)
Therefore,
Area(ABC) = (1/2)ABBC = (1/2)BC^2*tan(60)
sin(75)^2
=  ~= 1.07735
sin(60)

Posted by Bractals
on 20080211 13:41:41 