12.354. The base of the pyramid is an acute isosceles triangle, with the base equal to $a$ and the opposite angle equal to $\alpha$. The lateral edge of the pyramid, passing through the vertex of this angle, forms an angle $\beta$ with the plane of the base. Find the volume of the pyramid if the height of the pyramid passes through the point of intersection of the altitudes of the base.
Solution.

Let $EO$ be the height of the pyramid $EABC$, $AB=BC$, $AC=a$, $\angle ABC=\alpha$, $0^{\circ}<\alpha<90^{\circ}$, $\angle EBO=\beta$, $O$ is the point of intersection of the heights $BD$, $CK$, $AL$ - bases (Fig. 12.215).

Since $\triangle ABC$ is acute-angled, the point $O$ lies inside the triangle, and the points $L$ and $K$ lie on the segments $BC$ and $AB$.

In $\triangle ADB\left(\angle ADB=90^{\circ}\right)$:

$$
AB=\frac{AD}{\sin \angle ABD}=\frac{a}{2 \sin \frac{\alpha}{2}} ; BD=AD \operatorname{ctg} \angle ABD=\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}
$$

In $\triangle ALB\left(\angle ALB=90^{\circ}\right): BL=AB \cos \angle ABC=\frac{a \cos \alpha}{2 \sin \frac{\alpha}{2}}$.

In $\triangle OLB\left(\angle OLB=90^{\circ}\right): BO=\frac{BL}{\cos \angle OBL}=\frac{a \cos \alpha}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=a \operatorname{ctg} \alpha$.

In $\triangle EOB\left(\angle EOB=90^{\circ}\right): EO=OB \operatorname{tg} \angle EBO=a \operatorname{ctg} \alpha \operatorname{tg} \beta$.

$$
S_{\triangle ABC}=AD \cdot BD=\frac{a^{2}}{4} \operatorname{ctg} \frac{\alpha}{2}
$$

Volume of the pyramid:

$$
V=\frac{1}{3} S_{\triangle ABC} \cdot EO=\frac{1}{3} \cdot \frac{a^{2}}{4} \operatorname{ctg} \frac{\alpha}{2} \cdot a \operatorname{ctg} \alpha \operatorname{tg} \beta=\frac{a^{3}}{12} \operatorname{ctg} \alpha \operatorname{ctg} \frac{\alpha}{2} \operatorname{tg} \beta
$$

Answer: $\frac{a^{3}}{12} \operatorname{ctg} \alpha \operatorname{ctg} \frac{\alpha}{2} \operatorname{tg} \beta$.