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The 80-80-20 Triangle Problem, A Derivative, Solution #4
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ABC is an isosceles triangle with vertex angle  BAC = 20° and AB = AC. Point E is on AB such that AE = BC. Find the measure of AEC.
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Solution
Copyright © 1996-2008 Alexander Bogomolny
This is a trigonometric solution reported in [Leikin"].
Let E' be a point on AB with ACE' = 10°. We'll show that AE' = BC and, therefore, E = E'.
By the Law of Sines, in ΔACE',
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AE' / sin 10° = CE' / sin 20°.
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But, by the double angle formula, sin 20° = 2·sin 10° · cos 10°, so that
In ΔBCE',
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BC / sin 30° = CE' / sin 80° = CE' / cos 10°,
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so that
implying AE' = BC and, subsequently, E = E'. Thus ACE = 10° and AEC = 180° - 20° - 10° = 150°.
Reference
- R. Leikin, Dividable Triangles — What Are They?, Mathematics Teacher, May 2001, pp. 392–398.
Copyright © 1996-2008 Alexander Bogomolny
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