12.380. The base of a right prism is a right-angled triangle, one of whose acute angles is $\alpha$. The largest lateral face of the prism is a square. Find the angle between the intersecting diagonals of the other two lateral faces.
Solution.

Let $\triangle A B C$ be the base of the right prism $A B C A_{1} B_{1} C_{1}$ (Fig. 12.240) $\angle A C B=90^{\circ}, \angle B A C=\alpha$.

Since $A A_{1}=B B_{1}=C C_{1}, A B>A C$ and $A B>B C$, the face $A A_{1} B_{1} B$, which has the largest area, is a square.

Let the desired $\angle A_{1} C B_{1}=\beta, A B=A_{1} B_{1}=A A_{1}=1$.

In $\triangle A B C\left(\angle A C B=90^{\circ}\right)$ :

$A C=A B \cos \angle B A C=\cos \alpha ; B C=A B \sin \angle B A C=\sin \alpha$.

In $\triangle A_{1} A C\left(\angle A_{1} A C=90^{\circ}\right): A_{1} C=\sqrt{A A_{1}^{2}+A C^{2}}=\sqrt{1+\cos ^{2} \alpha}$.

In $\triangle B_{1} B C\left(\angle B_{1} B C=90^{\circ}\right): B_{1} C=\sqrt{B B_{1}{ }^{2}+B C^{2}}=\sqrt{1+\sin ^{2} \alpha}$.

From $\triangle A_{1} C B_{1}$ by the cosine rule:

$$
\begin{aligned}
& A_{1} B_{1}{ }^{2}=A_{1} C^{2}+B_{1} C^{2}-2 A_{1} C \cdot B_{1} C \cos \angle A_{1} C B_{1} \Rightarrow \\
& \Rightarrow 1=1+\cos ^{2} \alpha+1+\sin ^{2} \alpha-2 \sqrt{\left(1+\cos ^{2} \alpha\right)\left(1+\sin ^{2} \alpha\right)} \cdot \cos \beta \Leftrightarrow \\
& \Leftrightarrow 2 \sqrt{1+\cos ^{2} \alpha+\sin ^{2} \alpha+\cos ^{2} \alpha \sin ^{2} \alpha} \cdot \cos \beta=2 \Leftrightarrow \\
& \Leftrightarrow \sqrt{2+\frac{1}{4} \sin ^{2} 2 \alpha} \cdot \cos \beta=1 \Leftrightarrow \frac{\cos \beta \sqrt{8+\sin ^{2} 2 \alpha}}{2}=1 \Leftrightarrow \\
& \Leftrightarrow \beta=\arccos \frac{2}{\sqrt{8+\sin ^{2} 2 \alpha}}
\end{aligned}
$$

Answer: $\arccos \frac{2}{\sqrt{8+\sin ^{2} 2 \alpha}}$.