15. In an acute-angled triangle $A B C$, points $D, E$, and $F$ are the feet of the perpendiculars from $A, B$, and $C$ onto $B C, A C$ and $A B$, respectively. Suppose $\sin A=\frac{3}{5}$ and $B C=39$, find the length of $A H$, where $H$ is the intersection $A D$ with $B E$.
15. Ans: 52
First we know that $A E=A B \cos A$ and $A F=A C \cos A$. By cosine rule, $E F^{2}=$ $A E^{2}+A F^{2}-2 A E \times A F \cos A=\cos ^{2} A\left(A B^{2}+A C^{2}-2 A B \times A C \cos A\right)=$ $B C^{2} \cos ^{2} A$. Therefore $E F=B C \cos A$. It is easy to see that $A, E, H, F$ lie on a circle with diameter $A H$. Thus $A H=\frac{E F}{\sin A}=B C \cot A=39 \times \frac{4}{3}=52$.