3.41. The side of the larger base of a regular quadrilateral truncated pyramid is equal to $a$. The lateral edge and the diagonal of the pyramid form angles with the base plane that are equal to $\alpha$ and $\beta$, respectively. Find the area of the smaller base of the pyramid.
3.41. Given that $A B C D A_{1} B_{1} C_{1} D_{1}$ is a regular truncated quadrilateral pyramid, $A B=a, O O_{1} \perp(A B C), B_{1} E\left\|D_{1} F\right\| O O_{1}$, $\angle D_{1} D E=\alpha, \angle B_{1} D E=\beta$ (Fig. 3.43). Let $A_{1} B_{1}=x$; then $B_{1} D_{1}=x \sqrt{2}, B D=a \sqrt{2}$. We have

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-525.jpg?height=530&width=729&top_left_y=108&top_left_x=239)

Fig. 3.43

$$
\begin{aligned}
& F D=O D-O F=\frac{a \sqrt{2}}{2}-\frac{x \sqrt{2}}{2}=\frac{(a-x) \sqrt{2}}{2} \\
& E D=O D+O E=\frac{a \sqrt{2}}{2}+\frac{x \sqrt{2}}{2}=\frac{(a+x) \sqrt{2}}{2}
\end{aligned}
$$

From $\triangle D_{1} F D$ we find that $D_{1} F=F D \operatorname{tg} \alpha=\frac{(a-x) \sqrt{2}}{2} \operatorname{tg} \alpha$, and from $\Delta B_{1} E D$ - that $B_{1} E=E D \operatorname{tg} \beta=\frac{(a+x) \sqrt{2}}{2} \operatorname{tg} \beta$. Since $D_{1} F=B_{1} E=$ $=O O_{1}$, then

$$
(a-x) \operatorname{tg} \alpha=(a+x) \operatorname{tg} \beta, \text { or } x(\operatorname{tg} \beta+\operatorname{tg} \alpha)=a(\operatorname{tg} \alpha-\operatorname{tg} \beta) \text {, }
$$

i.e., $x=\frac{a \sin (\alpha-\beta)}{\sin (\alpha+\beta)}$. Therefore, $S_{A_{1} B_{1} C_{1} D_{1}}=\frac{a^{2} \sin ^{2}(\alpha-\beta)}{\sin ^{2}(\alpha+\beta)}$.