3.76. One of the sides of the base of a right triangular prism is equal to $a$, and the angles adjacent to it are $\alpha$ and $\beta$. Find the lateral surface area of the prism if its volume is $V$.
3.76. According to the condition, $A B C A B_{1} C_{1}$ is a straight prism, $A B=a, \angle B A C=\alpha$, $\angle A B C=\beta, V_{\text {mp }}=V$ (Fig. 3.77). Further, considering that $V=S_{\triangle A B C} H$, where the area of the base is expressed by the equality

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-548.jpg?height=456&width=470&top_left_y=403&top_left_x=686)

Fig. 3.77

$$
S_{\triangle A B C}=\frac{a^{2} \sin \alpha \sin \beta}{2 \sin \left(180^{\circ}-(\alpha+\beta)\right)}=\frac{a^{2} \sin \alpha \sin \beta}{2 \sin (\alpha+\beta)}
$$

we find $H=\frac{2 V \sin (\alpha+\beta)}{a^{2} \sin \alpha \sin \beta}$. According to the sine theorem, we have $\frac{B C}{\sin \alpha}=$ $=\frac{A C}{\sin \beta}=\frac{a}{\sin (\alpha+\beta)}$, from which $B C=\frac{a \sin \alpha}{\sin (\alpha+\beta)}, A C=\frac{a \sin \beta}{\sin (\alpha+\beta)}$. Therefore,

$$
P_{\triangle A B C}=\frac{a(\sin (\alpha+\beta)+\sin \alpha+\sin \beta)}{\sin (\alpha+\beta)}=
$$

$$
=\frac{2 a \sin \frac{\alpha+\beta}{2}\left(\cos \frac{\alpha+\beta}{2}+\cos \frac{\alpha-\beta}{2}\right)}{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}}=\frac{2 a \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{\cos \frac{\alpha+\beta}{2}} .
$$

$$
S_{\text {side }}=P_{\triangle A B C} H=\frac{2 a \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{\cos \frac{\alpha+\beta}{2}} \cdot \frac{2 V \sin (\alpha+\beta)}{a^{2} \sin \alpha \sin \beta}=\frac{2 V \sin \frac{\alpha+\beta}{2}}{a \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-549.jpg?height=507&width=492&top_left_y=385&top_left_x=46)

Fig. 3.78