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.

Draw a horizontal line from D to new point G on AC and another horizontal line from E to new point F on AB. The length of the side of the main triangle is now split into three lengths: BF, FD and DA. Assigning unit length to the base BC and using the Sine Rule, BF=EC=sin60/sin40, DB=sin70/sin30 so FD=sin70/sin30-sin60/sin40 and similarly DA=sin80/sin20-sin70/sin30. Next drop a vertical line down from D onto new point H on line FE. DH=FD*sin80. By similar triangles DE=DA/BA. So tan(DEF) = sin80*(sin70/sin30-sin60/sin40)/sin20*(sin80/sin20-sin70/sin30)
and CDE = 180-110-DEF. I made this come out to 13 degrees.

 

Posted by Jils on 2004-11-05 10:29:58