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
Iterative Solution by Charlie)
Charlie had an iterative solution that was correct. Then a number of others went off on tangents (sorry) and determined that there was no unique answer.
This is not an indeterminate problem. There is one, and only one, solution. It employs not-very-difficult trig.
PROOF:
Since the angle in question will be the same for all similar triangles, let BC, the base of the isosceles triangle, = 1
Therefore, AC and AB, the sides, = 1*sin 80/sin 20 = 2.879 (by the Law of Sines)
BD = sin 70/sin 30 = 1.879
CE = sin 60/sin 40 = 1.347
CD = sin 80/sin 30 = 1.970
AD = AB BD
AD = sin 80/sin 20 sin 70/sin 30 = 2.879 - 1.879 = 1.000
(interestingly, AD = BC, the triangle's base!)
AE = AC AE
AE = sin 80/sin 20 sin 60/sin 40 = 2.879 - 1.347 = 1.532
DE^2 = AE^2 + AD^2 2*AE*AD*cos 20 (Law of Cosines)
DE^2 = 1.532^2 + 1^2 2*1.532*1.000*0.940
DE^2 = 0.468
DE = 0.684
sin 20/DE = sin ADE/AE
sin ADE = AE*sin 20/DE
sin ADE = 1.532*0.342/0.684 = 0.766
angle CDE = 180 - angle BDE angle ADE
angle CDE = 180 - 30 arcsin 0.766
angle CDE = 150 130
angle CDE = 20
Charlie's answer was right.
There is only one solution
