12.384. The base of the pyramid is a right-angled triangle, with the hypotenuse equal to $c$, and the smaller of the acute angles equal to $\alpha$. The largest lateral edge forms an angle $\beta$ with the base plane. Find the volume of the pyramid if its height passes through the point of intersection of the medians of the base.
## Solution.

Let $EO$ be the height of the pyramid $EABC$ (Fig. 12.234), $O$ be the point of intersection of the medians of triangle $ABC, \angle ACB=90^{\circ}, AB=c, \angle BAC=\alpha, 0^{\circ}<\alpha<90^{\circ}$. Since $OB$ and $OA>OC$ $\Rightarrow EA$ is the largest lateral edge of the pyramid and $\angle EAO=\beta$.

In $\triangle ABC\left(\angle ACB=90^{\circ}\right): BC=c \sin \alpha ; AC=c \cos \alpha \Rightarrow$
$S_{\triangle ABC}=\frac{1}{2} BC \cdot AC=\frac{1}{2} c^{2} \sin \alpha \cos \alpha=\frac{1}{4} c^{2} \sin 2 \alpha ;$

$CD=\frac{1}{2} BC=\frac{1}{2} c \sin \alpha$.

In $\triangle ACD\left(\angle ACD=90^{\circ}\right): AD=\sqrt{AC^{2}+CD^{2}}=\sqrt{c^{2} \cos ^{2} \alpha+\frac{1}{4} c^{2} \sin ^{2} \alpha}=$ $=\frac{1}{2} c \sqrt{4 \cos ^{2} \alpha+\sin ^{2} \alpha}=\frac{1}{2} c \sqrt{1+3 \cos ^{2} \alpha}$.

$OA=\frac{2}{3} AD=\frac{1}{3} c \sqrt{1+3 \cos ^{2} \alpha}$.

In $\triangle AOE\left(\angle AOE=90^{\circ}\right): EO=AO \operatorname{tg} \angle EAO=\frac{1}{3} c \operatorname{tg} \beta \sqrt{1+3 \cos ^{2} \alpha}$.

The volume of the pyramid $V=\frac{1}{3} S_{\triangle ABC} \cdot EO=\frac{c^{3}}{36} \sin 2 \alpha \operatorname{tg} \beta \sqrt{1+3 \cos ^{2} \alpha}$.

Answer: $\frac{c^{3}}{36} \sin 2 \alpha \operatorname{tg} \beta \sqrt{1+3 \cos ^{2} \alpha}$.