XV OM - I - Problem 11

In triangle $ ABC $, angle $ A $ is $ 20^\circ $, $ AB = AC $. On sides $ AB $ and $ AC $, points $ D $ and $ E $ are chosen such that $ \measuredangle DCB = 60^\circ $ and $ \measuredangle EBC = 50^\circ $. Calculate the angle $ EDC $.
Let $ \measuredangle EDC = x $ (Fig. 9). Notice that $ \measuredangle ACB = \measuredangle  $ABC$  = 80^\circ $, $ \measuredangle CDB = 180^\circ-80^\circ-60^\circ = 40^\circ $, $ \measuredangle CEB = 180^\circ - 80^\circ-50^\circ = \measuredangle EBC $, hence $ EC = CB $. The ratio $ \frac{DC}{CE} $ of the sides of triangle $ EDC $ equals the ratio of the sides $ \frac{DC}{CB} $ of triangle $ BDC $, so the ratios of the sines of the angles opposite the corresponding sides in these triangles are equal:

The right side of the obtained equation can be transformed:

We need to find the convex angle $ x $ that satisfies the equation

or the equation

By transforming the products of sines into differences of cosines, we obtain an equivalent equation

Considering the condition $ 0 < x < 180^\circ $, we get $ x = 30^\circ $.
Note. The last part of the solution can be slightly shortened. Specifically, from the form of equation (1), it is immediately clear that it has a root $ x = 30^\circ $. No other convex angle satisfies this equation; if

then

so

thus, if $ 0 < x < 180^\circ $ and $ 0 < y < 180^\circ $, then $ x = y $.