12.325. The base of the pyramid is an acute isosceles triangle with a side length of $b$ and an angle at the base of $\alpha$. All lateral edges of the pyramid form the same angle $\beta$ with the base plane. Find the area of the section of the pyramid by a plane passing through the vertex of the given angle $\alpha$ and the height of the pyramid.
Solution.

Let $EO$ be the height of the pyramid $EABC$, $AC=AB=b$, $\angle ACB=\alpha$ (Fig. 12.186). Since all lateral edges form the same angle with the base plane, point $O$ is the center of the circle circumscribed around $\triangle ABC$, and since $\triangle ABC$ is isosceles and acute-angled, point $O$ lies on its height and bisector $AD$ (Fig. 12.186, a).

If $\triangle CEF$ is a section of the pyramid, then $S_{\triangle CEF}=\frac{1}{2} CF \cdot EO$.

In $\triangle ABC: \angle BAC=180^{\circ}-2\alpha, \angle OAC=90^{\circ}-\alpha$.

$OA=OC=R$ - the radius of the circle circumscribed around $\triangle ABC$.

Then $\angle ACO=\angle OAC=90^{\circ}-\alpha, \angle AFC=180^{\circ}-(\angle BAC+\angle ACO)=$ $=180^{\circ}-(180^{\circ}-2\alpha+90^{\circ}-\alpha)=3\alpha-90^{\circ}$.

In $\triangle AFC: \frac{FC}{\sin \angle FAC}=\frac{AC}{\sin \angle AFC} ; FC=\frac{b \sin (180^{\circ}-2\alpha)}{\sin (3\alpha-90^{\circ})}=-\frac{b \sin 2\alpha}{\cos 3\alpha}$.

In $\triangle ABC: OA=R=\frac{AC}{2 \sin \angle ABC}=\frac{b}{2 \sin \alpha}$.

In $\triangle AOE (\angle AOE=90^{\circ})$:

$$
\begin{aligned}
& EO=AO \operatorname{tg} \angle EAO=\frac{b \operatorname{tg} \beta}{2 \sin \alpha} \Rightarrow \\
& \Rightarrow S_{\triangle CEF}=\frac{1}{2} CF \cdot EO=\frac{1}{2} \cdot\left(-\frac{b \sin 2\alpha}{\cos 3\alpha}\right) \cdot \frac{b \operatorname{tg} \beta}{2 \sin \alpha}=-\frac{b^{2} \cos \alpha \operatorname{tg} \beta}{2 \cos 3\alpha} \\
& \text{Answer: }-\frac{b^{2} \cos \alpha \operatorname{tg} \beta}{2 \cos 3\alpha}
\end{aligned}
$$