10.196. Given a triangle $A B C$, in which $2 h_{c}=A B$ and $\angle A=75^{\circ}$. Find the measure of angle $C$.
## Solution.

Let the height $CD$ be denoted by $h$, and the segment $AD$ by $x$ (Fig. 10.6). We have $\angle ACD = 90^{\circ} - 75^{\circ} = 15^{\circ}$. Draw $AF$ such that $\angle CAF = \angle ACD = 15^{\circ}$. Then $\angle AFD = 30^{\circ}$, and from $\triangle ADF$ we get $AF = FC = 2x$, $DF = x \sqrt{3}$. But $DF = h - FC = h - 2x \Rightarrow x = \frac{h}{2 + \sqrt{3}} = h(2 - \sqrt{3})$. Since $AB = 2h$, we have $BD = AB - AD = 2h - x = h \sqrt{3}$. From $\triangle BDC$ we have $BC = \sqrt{BD^2 + CD^2} = 2h \Rightarrow \angle B = 30^{\circ}$ and $\angle C = 180^{\circ} - 75^{\circ} - 30^{\circ} = 75^{\circ}$.

Answer: $75^{\circ}$.