17. Given the lengths $h_{a}$ and $h_{b}$ of the altitudes of triangle $ABC$ dropped from vertices $A$ and $B$, and the length $l$ of the angle bisector of angle $C$. Find the angle $\hat{C}$.
17. Let $|A C|=b, \mid B C_{1}=a$. We have $b=\frac{h_{a}}{\sin \hat{C}}, a=$ $=\frac{h_{b}}{\sin \hat{C}}, l=\frac{2 a b \cos \frac{\hat{C}}{2}}{a+b}$ (see problem 17, section I). Therefore, $\boldsymbol{l}=\frac{2 h_{a} h_{b} \cos \frac{\hat{C}}{2}}{\sin \hat{C}\left(h_{a}+h_{b}\right)}$, from which $\sin \frac{\hat{C}}{2}=\frac{h_{a} h_{b}}{l\left(h_{a}+h_{b}\right)}$.