12.257. The base of the pyramid is a right-angled triangle inscribed in the base of the cone. The vertex of the pyramid coincides with the vertex of the cone. The lateral faces of the pyramid, containing the legs of the base, form angles $\alpha$ and $\beta$ with the plane of the base. Find the ratio of the volumes of the pyramid and the cone.
## Solution.

Let $DO$ be the common height of the given pyramid $DABC$ and the cone (Fig. 12.123). Since $O$ is the center of the circle circumscribed around the right triangle $ABC$, $O$ is the midpoint of the hypotenuse $AC$.

Draw perpendiculars $OE$ and $OF$ from point $O$ to the legs $AB$ and $BC$ in the plane of the base.

$OE$ and $OF$ are the projections of $DE$ and $DF$ on the base plane. Then $DE \perp AB$, $DF \perp BC$, $\angle DEO$ and $\angle DFO$ are the angles of inclination of the lateral faces $DAB$ and $BDC$ to the base plane, $\angle DEO=\beta, \angle DFO=\alpha$.

If $DO=a$, then from $\triangle DOE\left(\angle DOE=90^{\circ}\right): EO=DO \operatorname{ctg} \angle DEO=a \operatorname{ctg} \beta$.

From $\triangle DOF\left(\angle DOF=90^{\circ}\right): FO=DO \operatorname{ctg} \angle DFO=a \operatorname{ctg} \alpha$.

$AB=2 \cdot FO=2a \operatorname{ctg} \alpha, BC=2 \cdot EO=2a \operatorname{ctg} \beta$.

In $\triangle ABC\left(\angle ABC=90^{\circ}\right): AC=\sqrt{AB^{2}+BC^{2}}=2a \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}$.

The area of the base of the pyramid $S=\frac{1}{2} AB \cdot BC=2a^{2} \operatorname{ctg} \alpha \operatorname{ctg} \beta$, and its volume $V_{1}=\frac{1}{3} S \cdot DO=\frac{2}{3} a^{3} \operatorname{ctg} \alpha \operatorname{ctg} \beta$.

The radius of the base of the cone $R=\frac{1}{2} AC=a \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}$, its volume $V_{2}=\frac{1}{3} \pi R^{2} \cdot DO=\frac{1}{3} \pi a^{3}\left(\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta\right)$.

Thus, $\frac{V_{1}}{V_{2}}=\frac{2 \operatorname{ctg} \alpha \operatorname{ctg} \beta}{\pi\left(\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta\right)}$.

Answer: $\frac{2 \operatorname{ctg} \alpha \operatorname{ctg} \beta}{\pi\left(\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta\right)}$.