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 Measure that angle IV (Posted on 2004-11-02)
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.

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

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 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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