12.297. In a regular triangular prism, the side of the base is equal to $a$, and the angle between the non-intersecting diagonals of two lateral faces is $\alpha$. Find the height of the prism.
## Solution.

Let's complete the regular prism $A B C A_{1} B_{1} C_{1}$ to a parallelepiped $A D B C A_{1} D_{1} B_{1} C_{1}$ (Fig. 12.159). Since $B D_{1} \| C A_{1}$, the angle between the skew lines $A_{1} C$ and $B C_{1}$ is $\angle C_{1} B D_{1}=\alpha$.

Since $\Delta A_{1} B_{1} C_{1}$ is equilateral, $A_{1} D_{1} B_{1} C_{1}$ is a rhombus, $C_{1} O_{1}$ is the height of $\Delta A_{1} B_{1} C_{1}$, $C_{1} O_{1}=\frac{a \sqrt{3}}{2}$, and $C_{1} D_{1}=a \sqrt{3}$.

Let $B C_{1}=B D_{1}=x$.

In $\triangle C_{1} B D_{1}$: $C_{1} D_{1}^{2}=B C_{1}^{2}+B D_{1}^{2}-2 B C_{1} \cdot B D_{1} \cos \angle C_{1} B D_{1}$.

We have $3 a^{2}=2 x^{2}-2 x^{2} \cos \alpha \Leftrightarrow 4 x^{2} \sin ^{2} \frac{\alpha}{2}=3 a^{2} \Leftrightarrow x^{2}=\frac{3 a^{2}}{4 \sin ^{2} \frac{\alpha}{2}}$.

In $\triangle B B_{1} C_{1}\left(\angle B B_{1} C_{1}=\frac{\pi}{2}\right)$:

$B B_{1}=\sqrt{B C_{1}^{2}-B_{1} C_{1}^{2}}=\sqrt{\frac{3 a^{2}}{4 \sin ^{2} \frac{\alpha}{2}}-a^{2}}=\frac{a}{\sin \frac{\alpha}{2}} \sqrt{\frac{3}{4}-\sin ^{2} \frac{\alpha}{2}}=$

$$
\begin{aligned}
& =\frac{a}{\sin \frac{\alpha}{2}} \sqrt{\sin ^{2} \frac{\pi}{3}-\sin ^{2} \frac{\alpha}{2}}=\frac{a}{\sin \frac{\alpha}{2}} \sqrt{\frac{1-\cos \frac{2 \pi}{3}}{2}-\frac{1-\cos \alpha}{2}}= \\
& =\frac{a}{\sin \frac{\alpha}{2}} \sqrt{\frac{\cos \alpha-\cos \frac{2 \pi}{3}}{2}}=\frac{a \sqrt{\sin \left(\frac{\pi}{3}-\frac{\alpha}{2}\right) \sin \left(\frac{\pi}{3}+\frac{\alpha}{2}\right)}}{\sin \frac{\alpha}{2}} \\
& \text { Therefore: } \frac{a \sqrt{\sin \left(\frac{\pi}{3}-\frac{\alpha}{2}\right) \sin \left(\frac{\pi}{3}+\frac{\alpha}{2}\right)}}{\sin \frac{\alpha}{2}}
\end{aligned}
$$