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Measure that angle IV (Posted on 2004-11-02) Difficulty: 4 of 5
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.

See The Solution Submitted by Brian Smith    
Rating: 3.6667 (9 votes)

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Some Thoughts Part II, but still not finished re: Part of solution | Comment 4 of 21 |
(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
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