12.402 In a triangle, two sides $a$ and $b(a>b)$ and the area $S$ are given. Find the angle between the height and the median drawn to the third side.
## Solution.

$$
\begin{aligned}
& \text { Let in the given } \triangle A B C \quad a=B C, \dot{b}=A B, B D \text { - altitude, } \\
& B M-\text { median of } \triangle A B C . \text { Then } \\
& B D^{2}=A B^{2}-\left(\frac{A C}{2}-D M\right)^{2}=B C^{2}-\left(\frac{A C}{2}+D M\right)^{2} \Rightarrow \\
& \Rightarrow a^{2}-b^{2}=2 \cdot A C \cdot D M \Rightarrow a^{2}-b^{2}=2 \cdot A C \cdot B D \cdot \operatorname{tg} \angle D B M=4 S \cdot \operatorname{tg} \angle D B M \Rightarrow \\
& \Rightarrow \angle D B M=\operatorname{arctg}\left(\frac{a^{2}-b^{2}}{4 S}\right)
\end{aligned}
$$

Answer: $\operatorname{arctg} \frac{a^{2}-b^{2}}{4 S}$.