12.436 The base of the prism is a parallelogram with an acute angle $\alpha$. The lateral edge passing through the vertex of this angle $\alpha$ is equal to $b$ and forms angles, each equal to $\boldsymbol{\beta}$, with the adjacent sides of the base. Find the height of the prism.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-309.jpg?height=320&width=361&top_left_y=1301&top_left_x=177)

Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be the given prism, $A A_{1}=b, \angle B A D=\alpha, \angle A_{1} A B=\angle A_{1} A D=\beta, \quad A_{1} O$ be the height of the prism.

Drop perpendiculars $A_{1} K$ and $A_{1} M$ to the edges $A D$ and $A B$. Then $O K \perp A D, O M \perp A B$ (Theorem 12). $\Delta A_{1} A K=\Delta A_{1} A M \Rightarrow A K=A M \Rightarrow A O$ is the bisector of angle $B A D \Rightarrow$
$A K=b \cdot \cos \beta, A O=\frac{A K}{\cos \frac{\alpha}{2}}=\frac{b \cos \beta}{\cos \frac{\alpha}{2}} \Rightarrow A_{1} O=\sqrt{A A_{1}^{2}-A O^{2}}=$

$=\frac{b}{\cos \frac{\alpha}{2}} \sqrt{\cos ^{2} \frac{\alpha}{2}-\cos ^{2} \beta}=\frac{b}{\cos \frac{\alpha}{2}} \sqrt{\sin \left(\beta-\frac{\alpha}{2}\right) \cdot \sin \left(\beta+\frac{\alpha}{2}\right)}$

Answer: $\frac{b}{\cos \frac{\alpha}{2}} \sqrt{\sin \left(\beta+\frac{\alpha}{2}\right) \cdot \sin \left(\beta-\frac{\alpha}{2}\right)}$