Triangle ABC is isosceles with AB=AC. Point D is on side AB such that angle BCD is 70 degrees. Point E is on side AC such that angle EBC is 60 degrees. Angle ABE equals 20 degrees, and angle DCE equals 10 degrees.

Find angle EDC. Justify your answer.

(In reply to Part of solution by Dustin)

Ok, BCsin(60) = ECsin(40) => BC = ECsin(40)/sin(60)

BCsin(70) = BDsin(30) => BC = BDsin(30)/sin(70)

So ECsin(40)/sin(60) = BDsin(30)/sin(70)

So ECsin(40)sin(70) = BDsin(30)sin(60).

So EC = BDsin(30)sin(60)/ (sin(40)sin(70))

sin(20)/DE = sin(130-x)/BD, and sin(x)/EC = sin(10)/DE

DEsin(130-x) = BDsin(20), and DEsinx = ECsin(10)

DEsinx = BDsin(30)sin(60)sin(10)/(sin(40)sin(70))

DE = BDsin(30)sin(60)sin(10)/(sin(40)sin(70)sinx)

BDsin(30)sin(60)sin(10)sin(130-x)/(sin(40)sin(70)sinx) = BDsin(20)

sin(30)sin(60)sin(10)sin(130-x) = sin(20)sin(40)sin(70)sinx

There must be some way to solve for x, but I can't see it.

Posted by Dustin on 2004-11-02 19:18:08