3.77. The base of a right prism $A B C A_{1} B_{1} C_{1}\left(A A_{1}\left\|B B_{1}\right\| C C_{1}\right)$ is an isosceles triangle $A B C(A B=A C)$, with a perimeter of $2 p$ and the angle at vertex $A$ equal to $\alpha$. A plane is drawn through side $B C$ and vertex $A_{1}$, forming an angle $\beta$ with the base plane. Find the volume of the prism.
3.77. Given that $A B C A_{1} B_{1} C_{1}$ is a right prism, $A B=A C, A B+A C+$ $+B C=2 p, \angle B A C=\alpha, A_{1} B C$ is a section of the prism, $\angle A_{1} B C A=\beta$ (Fig. 3.78). Draw $A_{1} K \perp B C$ and let $A A_{1}=h$; then from $\triangle A_{1} A K$ we find $A K=h \operatorname{ctg} \beta$, and from $\triangle A K B$ we get

$$
A B=\frac{A K}{\cos \frac{\alpha}{2}}=\frac{h \operatorname{ctg} \beta}{\cos \frac{\alpha}{2}}
$$

$B K=h \operatorname{ctg} \beta \operatorname{tg} \frac{\alpha}{2}$. Therefore,

$B C=2 h \operatorname{ctg} \beta \operatorname{tg} \frac{\alpha}{2}$, from which

$$
2 p=\frac{2 h \operatorname{ctg} \beta}{\cos \frac{\alpha}{2}}+2 h \operatorname{ctg} \beta \operatorname{tg} \frac{\alpha}{2}, \text { or } p \cos \frac{\alpha}{2}=h \operatorname{ctg} \beta\left(1+\sin \frac{\alpha}{2}\right) \text {, }
$$

i.e.

$$
h=\frac{p \cos \frac{\alpha}{2} \operatorname{tg} \beta}{1+\sin \frac{\alpha}{2}}=\frac{p \sin \left(\frac{\pi-\alpha}{2}\right) \operatorname{tg} \beta}{1+\cos \left(\frac{\pi-\alpha}{2}\right)}=p \operatorname{tg} \frac{\pi-\alpha}{4} \operatorname{tg} \beta
$$

Thus,

$$
\begin{aligned}
& V=S_{\triangle A B C} \cdot A A_{1}=\frac{1}{2} B C \cdot A K \cdot A A_{1}=\frac{1}{2} 2 h^{3} \operatorname{ctg}^{2} \beta \operatorname{tg} \frac{\alpha}{2}= \\
& =p^{3} \operatorname{tg}^{3} \frac{\pi-\alpha}{4} \operatorname{tg}^{3} \beta \operatorname{ctg}^{2} \beta \operatorname{tg} \frac{\alpha}{2}=p^{3} \operatorname{tg}^{3} \frac{\pi-\alpha}{4} \operatorname{tg} \beta \operatorname{tg} \frac{\alpha}{2}
\end{aligned}
$$