3.78. The base of a right prism is an equilateral triangle. A plane is drawn through one of its sides, cutting off a pyramid from the prism, the volume of which is equal to $V$. Find the area of the section, if the angle between the cutting plane and the base plane is $\alpha$.
3.78. By the condition, $A B C A_{1} B_{1} C_{1}$ is a regular triangular prism, $S A B C$ is a pyramid, $V_{S A B C}=V,((S A B);(A B C))=$ $=\alpha$ (Fig. 3.79). Let $A B=a, S C=h$; then $V=\frac{1}{3} \frac{a^{2} \sqrt{3}}{4} h, S_{\triangle S A B}=\frac{a^{2} \sqrt{3}}{4 \cos \alpha}$ and $h=\frac{a \sqrt{3}}{2} \operatorname{tg} \alpha$. Therefore, $V=\frac{a^{3} \operatorname{tg} \alpha}{8}$, hence

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-550.jpg?height=398&width=425&top_left_y=105&top_left_x=720)

Fig. 3.79 where $a=\frac{2 \sqrt[3]{V}}{\sqrt[3]{\operatorname{tg} \alpha}}$. Finally, we find

$$
S_{\triangle S A B}=\frac{\sqrt{3} \sqrt[3]{V^{2}}}{\cos \alpha^{\sqrt[3]{\operatorname{tg}^{2} \alpha}}}=\sqrt[3]{\frac{3 \sqrt{3} V^{2}}{\sin ^{2} \alpha \cos \alpha}}
$$