3.97. A cylinder is inscribed in a cone; the lower base of the cylinder lies in the plane of the base of the cone. A line passing through the center of the upper base of the cylinder and a point on the circumference of the base of the cone makes an angle $\alpha$ with the plane of the base. Find the ratio of the volumes of the cone and the cylinder, if the angle between the generatrix and the height of the cone is $\beta$.
3.97. Let $PMN$ be the axial section of a cone, $KLN_1M_1$ be the axial section of an inscribed cylinder, $\angle O_1MK=\alpha$, $PO$ be the height of the cone, and $\angle MPO=\beta$ (Fig. 3.97); we need to find $V_{\text{cone}}: V_{\text{cyl}}$. Let the radius of the base of the cone be $R$; then $OP=R \operatorname{ctg} \beta$ (from $\triangle POM$) and $V_{\text{cone}}=\frac{1}{3} \pi R^{3} \operatorname{ctg} \beta$. From $\Delta O_1OM$ we find $OO_1=R \operatorname{tg} \alpha$. Since $M_1K=OO_1$, from $\Delta M_1MK$, where $\angle MM_1K=\angle MPO=\beta$, we get $MK=R \operatorname{tg} \alpha \operatorname{tg} \beta$. Therefore,

$$
OK=OM-MK=R-R \operatorname{tg} \alpha \operatorname{tg} \beta=\frac{R \cos (\alpha+\beta)}{\cos \alpha \cos \beta}
$$

from which

$$
\begin{gathered}
V_{\text{cyl}}=\pi OK^{2} \cdot OO_1=\frac{\pi R^{3} \cos^{2}(\alpha+\beta) \operatorname{tg} \alpha}{\cos^{2} \alpha \cos^{2} \beta} \\
\frac{V_{\text{cone}}}{V_{\text{cyl}}}=\frac{\pi R^{3} \operatorname{ctg} \beta \cos^{2} \beta \cos^{2} \alpha}{3 \pi R^{3} \cos^{2}(\alpha+\beta) \operatorname{tg} \alpha}=\frac{\cos^{3} \alpha \cos^{3} \beta}{3 \sin \alpha \sin \beta \cos^{2}(\alpha+\beta)} .
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-563.jpg?height=396&width=587&top_left_y=1396&top_left_x=45)

Fig. 3.98