### 12.445 The base of an oblique prism is a right-angled triangle with an acute angle $\alpha$. The lateral face containing the hypotenuse is perpendicular to the base, while the lateral face containing the adjacent leg to the given angle forms an acute angle $\beta$ with the base. Find the acute angle between the third lateral face and the base.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-315.jpg?height=374&width=393&top_left_y=1205&top_left_x=220)

Let $A B C A_{1} B_{1} C_{1}$ be the given prism. $A B B_{1} A_{1} \perp A B C, \angle B C A=90^{\circ}, \angle C B A=\alpha$. Let $B_{1} L$ and $A_{1} K$ be the heights of the prism.

By the condition, they belong to the face $A B B_{1} A_{1}$. Drop perpendiculars $A_{1} M$ and $B_{1} N$ to the edges $A C$ and $B C$ respectively. Then $K M \perp C M, N L \perp B C$ (Theorem 12) $\Rightarrow \angle B_{1} N L$ and $\angle A_{1} M K$ are the angles between the corresponding faces and the base $A B C \Rightarrow \quad \angle B_{1} N L=\beta . B L=A K \quad$ (projections of equal obliques $\left.B_{1} B, A_{1} A\right) \Rightarrow \triangle A K M=\triangle B N L$

$$
\begin{gathered}
(\angle A K M=\angle C B A=\alpha) \Rightarrow A M=N L \Rightarrow \operatorname{tg} \alpha=\frac{A M}{K M}=\frac{N L}{K M}, \operatorname{tg} \beta=\frac{B_{1} L}{N L}=\frac{A_{1} K}{N L} \Rightarrow \\
\Rightarrow \operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{A_{1} K}{K M}=\operatorname{tg} \angle A_{1} M K
\end{gathered}
$$

Answer: $\angle A_{1} M K=\operatorname{arctg}(\operatorname{tg} \alpha \cdot \operatorname{tg} \beta)$