12.253. 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 $\beta$.
## Solution.

Let $O$ be the center of the bases of the given cone and the cylinder inscribed in it, $O_{1}$ be the center of the upper base of the cylinder, $SA$ be the generatrix of the cone, $CB$ be the generatrix of the cylinder, $\angle ASO=\beta, \angle O_{1}AO=\alpha$, (Fig. 12.119).

If the radius of the base of the cone $OA=1$, then from $\triangle AOS\left(\angle AOS=90^{\circ}\right)$: $SO=OA \operatorname{ctg} \angle ASO=\operatorname{ctg} \beta$.

The volume of the cone $V_{1}=\frac{1}{3} \pi \cdot OA^{2} \cdot SO=\frac{1}{3} \pi \operatorname{ctg} \beta$.

From $\triangle O_{1}OA\left(\angle O_{1}OA=90^{\circ}\right): O_{1}O=OA \operatorname{tg} \angle O_{1}AO=\operatorname{tg} \alpha$.

$BC=OO_{1}=\operatorname{tg} \alpha$.

$\angle ACB=\angle ASO=\beta$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1005.jpg?height=785&width=761&top_left_y=76&top_left_x=306)

Fig. 12.120

From $\triangle ABC\left(\angle ABC=90^{\circ}\right): AB=BC \operatorname{tg} \angle ACB=\operatorname{tg} \alpha \operatorname{tg} \beta$.

$OB=OA-AB=1-\operatorname{tg} \alpha \operatorname{tg} \beta=\frac{\cos (\alpha+\beta)}{\cos \alpha \cos \beta}$.

The volume of the cylinder $V_{2}=\pi \cdot OB^{2} \cdot OO_{1}=\frac{\pi \cos^{2}(\alpha+\beta) \operatorname{tg} \alpha}{\cos^{2} \alpha \cos^{2} \beta}$.

Then $\frac{V_{1}}{V_{2}}=\frac{\operatorname{ctg} \beta \cos^{2} \alpha \cos^{2} \beta}{3 \cos^{2}(\alpha+\beta) \operatorname{tg} \alpha}=\frac{\cos^{3} \alpha \cos^{3} \beta}{3 \sin \alpha \sin \beta \cos^{2}(\alpha+\beta)}$.

Answer: $\frac{\cos^{3} \alpha \cos^{3} \beta}{3 \sin \alpha \sin \beta \cos^{2}(\alpha+\beta)}$.