12.399 In a parallelogram, the two sides are $\boldsymbol{a}$ and $\boldsymbol{b}(\boldsymbol{a}>\boldsymbol{b})$ and the height $h$, drawn to the larger side. Find the acute angle between the diagonals of the parallelogram.
## Solution.

$$
\begin{aligned}
& \text { } \\
& \text { Let } ABCD \text { be the given parallelogram, } \\
& =a^{2}-\left(\frac{AC}{2}+OK\right)^{2} \Rightarrow 2 \cdot AC \cdot OK=a^{2}-b^{2} \Rightarrow \\
& \Rightarrow S_{ABCD}=AC \cdot BK=AC \cdot OK \cdot \operatorname{tg} \alpha=\frac{a^{2}-b^{2}}{2} \cdot \operatorname{tg} \alpha
\end{aligned}
$$

On the other hand, $S_{ABCD}=a \cdot h \Rightarrow \operatorname{tg} \alpha=\frac{2 a h}{a^{2}-b^{2}}$.

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