12.435 The base of the prism is an equilateral triangle with side $a$. The lateral edge is equal to $b$ and forms angles $\alpha$ and $\beta$ with the intersecting sides of the base. Find the volume of the prism.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-309.jpg?height=311&width=360&top_left_y=262&top_left_x=163)

Let $A_{1} D$ be the height of the given prism $A B C A_{1} B_{1} C_{1}$, where $\triangle A B C$ is equilateral, $\angle A_{1} A B = \alpha$, and $\angle A_{1} A C = \beta$. Drop perpendiculars $A_{1} N$ and $A_{1} M$ to the edges $A B$ and $A C$. Then $D M \perp A C$ and $D N \perp A B$ (Theorem 12).

Extend $M \dot{D}$ to intersect $A B$ at point $K$.

Then $A M = b \cos \beta$, $A N = b \cdot \cos \alpha$, $A K = \frac{A M}{\cos 60^{\circ}} = 2 b \cos \beta$.

$N K = A K - A N = b(2 \cos \beta - \cos \alpha) \Rightarrow D K = \frac{N K}{\cos 30^{\circ}} = \frac{2 b}{\sqrt{3}}(2 \cos \beta - \cos \alpha)$.

By the cosine rule for $\Delta A A_{1} K: A_{1} K^{2} = b^{2} + 4 b^{2} \cos^{2} \beta - 4 b^{2} \cos \alpha \cdot \cos \beta$.

Thus, $A_{1} D^{2} = A_{1} K^{2} - D K^{2} = \frac{b^{2}}{3}\left[3 - 4\left(\cos^{2} \alpha - \cos \alpha \cdot \cos \beta + \cos^{2} \beta\right)\right]$ and the volume of the prism is $\frac{a^{2} \sqrt{3}}{4} \cdot A_{1} D$.

Answer: $\frac{a^{2} b}{4} \sqrt{3 - 4\left(\cos^{2} \alpha - \cos \alpha \cdot \cos \beta + \cos^{2} \beta\right)}$.