3.80. The base of the pyramid is an isosceles trapezoid, with the lateral side equal to \(a\), and the acute angle equal to \(\alpha\). All lateral faces form the same angle \(\beta\) with the base of the pyramid. Find the total surface area of the pyramid.
3.80. By the condition, $S A B C D$ is a pyramid, $A B \| D C, A D=B C, A D=a$, $\angle A D C=\alpha, \angle S A B C=\angle S B C D=$ $=\angle S D C A=\angle S A D B=\beta$ (Fig. 3.81). Construct the linear angles of the dihedral angles at the base. Draw $S O \perp(A B C)$ and $O E, O F, O G, O H$ - perpendiculars to the sides of the base. Since $\triangle S O E=\triangle S O F=\Delta S O Q=$ $=\triangle S O H$, then $O E=O F=O Q=O H$, i.e., $O$ is the center of the circle inscribed in the base $A B C D$; then $A B+D C=$ $=2 A D=2 a$. Draw $A M \perp D C$ and from

$\triangle A M D$ find $A M=a \sin \alpha$. Further, we have

$$
\begin{gathered}
S_{A B C D}=\frac{A B+D C}{2} \cdot A M=a^{2} \sin \alpha ; S_{\text {side }}=\frac{a^{2} \sin \alpha}{\cos \beta} \\
S_{\text {full }}=\frac{a^{2} \sin \alpha}{\cos \beta}+a^{2} \sin \alpha=\frac{a^{2} \sin \alpha(1+\cos \beta)}{\cos \beta}=\frac{2 a^{2} \sin \alpha \cos ^{2} \frac{\beta}{2}}{\cos \beta} .
\end{gathered}
$$