12.218. The base of a right prism is a triangle with side $a$ and adjacent angles $\alpha$ and $\beta$. A plane is drawn through the side of the base at an angle $\varphi$ to it, intersecting the opposite lateral edge. Find the volume of the resulting triangular pyramid.
## Solution.

Let $\triangle A B C$ be the base of the right prism $A B C A_{l} B_{1} C_{1}$ (Fig.12.85), $\angle A C B=\alpha, \angle A B C=\beta, B C=a, E$ be the point of intersection of the section of the prism passing through $B C$ with its lateral edge $A A_{1}$. Then $A E$ is the height

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0967.jpg?height=508&width=680&top_left_y=198&top_left_x=92)

Fig. 12.85

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0967.jpg?height=636&width=445&top_left_y=29&top_left_x=815)

Fig. 12.86

of the obtained pyramid and its volume $V=\frac{1}{3} S_{\triangle A B C} \cdot E A, A F$ is the height of $\triangle A B C$ and the projection $E F$ on the plane $A B C$. Thus $E F \perp B C$ and $\angle E F A=\varphi$. In $\triangle A B C: \angle B A C=180^{\circ}-(\alpha+\beta)$

\[
\begin{aligned}
& A B=\frac{C B \sin \angle A C B}{\sin \angle B A C}=\frac{a \sin \alpha}{\sin (\alpha+\beta)} \Rightarrow S_{\triangle A B C}=\frac{1}{2} A B \cdot B C \sin \angle A B C= \\
& =\frac{a^{2} \sin \alpha \sin \beta}{2 \sin (\alpha+\beta)}
\end{aligned}
\]

Since $S_{\triangle A B C}=\frac{1}{2} B C \cdot A F$, then $A F=\frac{2 S_{\triangle A B C}}{B C}=\frac{a \sin \alpha \sin \beta}{\sin (\alpha+\beta)}$.

In $\triangle E A F\left(\angle E A F=90^{\circ}\right): E A=A F \operatorname{tg} \angle E F A=\frac{a \sin \alpha \sin \beta \operatorname{tg} \varphi}{\sin (\alpha+\beta)}$.

Therefore, $V=\frac{1}{3} \cdot \frac{a^{2} \sin \alpha \sin \beta}{2 \sin (\alpha+\beta)} \cdot \frac{a \sin \alpha \sin \beta \operatorname{tg} \varphi}{\sin (\alpha+\beta)}=\frac{a^{3} \sin ^{2} \alpha \sin ^{2} \beta \operatorname{tg} \varphi}{6 \sin ^{2}(\alpha+\beta)}$.

Answer: $\frac{a^{3} \sin ^{2} \alpha \sin ^{2} \beta \operatorname{tg} \varphi}{6 \sin ^{2}(\alpha+\beta)}$.