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 re: Not sure why I don't get the same answer by Charlie)

Yes, Sorry, well spotted - I meant DG!  I noticed a problem with my solution (as DG is not equal to HE). Never mind  as HE = DG+FH and FH/y = cos 80, so.....

tan DEH = [sin80*(sin70/sin30 - sin60/sin40)]/[(sin80/sin20)(sin80/sin20-sin70/sin30) + cos80*(sin70/sin30-sin60/sin40)]

DEH = 39.24 degrees

and EDC = 30.76 degrees.

 

 

 

Edited on November 11, 2004, 9:44 am

Posted by Jils on 2004-11-11 08:43:30