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',
AE' / sin 10° = CE' / sin 20°.
But, by the double angle formula, sin 20° = 2·sin 10° · cos 10°, so that
CE' = 2·AE'·cos 10°.
In ΔBCE',
BC / sin 30° = CE' / sin 80° = CE' / cos 10°,
so that
CE' = 2·BC·cos 10°,
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.
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