3.83. The height of a regular triangular pyramid is $H$. The lateral face forms an angle $\alpha$ with the base plane. A plane is drawn through a side of the base and the midpoint of the opposite lateral edge. Find the area of the resulting section.
3.83. Given that $S A B C$ is a regular triangular pyramid, $S O \perp (A B C)$, $S O=H$, $\angle S A C B=\alpha$, $S M=M B$, and $A M C$ is a section (Fig. 3.84). Draw $S K \perp A C$; then $O K \perp A C$ and $\angle S K O=\alpha$ as the linear angle of the dihedral angle $A C$. From $\triangle S O K$ we find $O K=H$ ctg $\alpha$, $A C=2 \sqrt{3} O K=2 H \sqrt{3} \operatorname{ctg} \alpha$. Draw $M M_{1} \| S O$; then $M M_{1}=\frac{H}{2}$, $K M_{1}=2 H \operatorname{ctg} \alpha$. From $\triangle M K M_{1}$ we find

$$
M K=\sqrt{\frac{H^{2}}{4}+4 H^{2} \operatorname{ctg}^{2} \alpha}=\frac{H}{2} \sqrt{1+16 \operatorname{ctg}^{2} \alpha}
$$

Thus,

$$
\begin{gathered}
S_{\triangle M A C}=\frac{1}{2} A C \cdot M K=H \sqrt{3} \operatorname{ctg} \alpha \cdot \frac{H}{2} \sqrt{1+16 \operatorname{ctg}^{2} \alpha}= \\
=\frac{H^{2} \sqrt{3} \operatorname{ctg} \alpha}{2} \sqrt{1+16 \operatorname{ctg}^{2} \alpha}
\end{gathered}
$$