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.

The knowledge that a triangle's three angles add up to 180 degrees will get you this far: (P.S. All my measurements are in degrees)

ABC = 80, ABE = 20, ACB = 80, ACD = 10, ADC = 150, AEB = 140, ADE = 150-x, AED = x + 10

BAC = 20, BCD = 70, BDC = 30, BEC = 40

CDE = x, CED = 170-x, BED = 130-x.

If you call the intersection of BE and DC F,

BFD and CFE = 130, and BFC and DFE = 50.

sin(60)/EC = sin(40)/BC, and sin(30)/BC = sin(70)/BD

BCsin(60) = ECsin(40), and BDsin(30) = BCsin(70)

I don't know what sin(40) and sin(70) are, but I think you have to use the lengths of BD and EC to find DE or something like that.   I'm going to look at the other posts now.

Edited on November 2, 2004, 6:57 pm

Posted by Dustin on 2004-11-02 18:52:30