12.362. The base of the pyramid is a rhombus with side $a$. Two lateral faces of the pyramid are perpendicular to the base plane and form an angle $\beta$ between them. The other two lateral faces form an angle $\alpha$ with the base plane. Find the lateral surface area of the pyramid.
Solution.

Rhombus $ABCD$ is the base of the pyramid $EABCD$ (Fig. 12.222), with faces $ABE$ and $CBE$ perpendicular to the base plane, making their common edge $EB$ the height of the pyramid.

Then $EB \perp AB, EB \perp BC, \angle ABC$ is the angle between the faces $ABE$ and $CBE$, $\triangle ABC = \beta$.

$BK$ is the height of rhombus $ABCD$. Then $EK \perp AD$ and $\angle EKB$ is the angle of inclination of face $AED$ to the base plane, $\angle EKB = \alpha$.

$\triangle ABD$ is the projection of $\triangle AED$ onto the base plane.

Then $S_{\triangle AED} = S_{\triangle CED} = \frac{S_{\triangle ABD}}{\cos \alpha}$;

$2 S_{\triangle AED} = \frac{S_{ABCD}}{\cos \alpha} = \frac{a^2 \sin \beta}{\cos \alpha}$.

$BK = a \sin \beta$.

From $\triangle EBK (\angle EBK = 90^\circ): EB = BK \operatorname{tg} \angle EKB = a \sin \beta \operatorname{tg} \alpha$.

$$
S_{\triangle ABE} = \frac{1}{2} AB \cdot EB = \frac{1}{2} a^2 \sin \beta \operatorname{tg} \alpha
$$

Lateral surface area of the pyramid:

$$
\begin{aligned}
& S = 2 S_{\triangle ABE} + 2 S_{\triangle AED} = a^2 \sin \beta \operatorname{tg} \alpha + \frac{a^2 \sin \beta}{\cos \alpha} = \frac{a^2 \sin \beta (1 + \sin \alpha)}{\cos \alpha} = \\
& = \frac{a^2 \sin \beta \left(1 + \cos \left(\frac{\pi}{2} - \alpha\right)\right)}{\cos \alpha} = \frac{2 a^2 \sin \beta \cos^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)}{\cos \alpha} \\
& \text{Answer: } \frac{2 a^2 \sin \beta \cos^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)}{\cos \alpha}
\end{aligned}
$$