12.119. A sphere is inscribed in a cone. The radius of the circle where the cone and the sphere touch is $r$. Find the volume of the cone if the angle between the height and the slant height of the cone is $\alpha$.
## Solution.

The volume of the cone is: $V=\frac{1}{3} \pi R_{\mathrm{k}}^{2} \cdot H$ (Fig. 12.119). From $\triangle A O_{1} O$ $\left(\angle A O_{1} O=90^{\circ}\right) O A=R_{\text {m }}=\frac{r}{\cos \alpha} ; O O_{1}=r \operatorname{tg} \alpha$. From $\triangle S O_{1} A$ $\left(\angle S O_{1} A=90^{\circ}\right) S O_{1}=\frac{r}{\operatorname{tg} \alpha}$. The height of the cone:

$$
\begin{aligned}
& H=S O_{1}+O_{1} O+R_{\mathrm{m}}=r\left(\frac{1}{\cos \alpha}+\operatorname{tg} \alpha+\frac{1}{\operatorname{tg} \alpha}\right)= \\
& =r\left(\frac{(1+\sin \alpha) \sin \alpha+\cos ^{2} \alpha}{\cos \alpha \sin \alpha}\right)=2 r\left(\frac{\sin \alpha+1}{\sin 2 \alpha}\right)
\end{aligned}
$$

From $\triangle S B C(\angle S B C)$

$$
R_{\mathrm{k}}=H \operatorname{tg} \alpha=r\left(\frac{\sin \alpha+1}{\sin \alpha \cos \alpha}\right) \cdot \frac{\sin \alpha}{\cos \alpha}=r\left(\frac{\sin \alpha+1}{\cos ^{2} \alpha}\right) . \text { Then }
$$

$$
\begin{aligned}
& V=\frac{1}{3} \pi r^{2}\left(\frac{\sin \alpha+1}{\cos ^{2} \alpha}\right)^{2} \cdot 2 r \cdot\left(\frac{\sin \alpha+1}{\sin 2 \alpha}\right)=\frac{1}{3} \pi r^{3} \frac{(\sin \alpha+1)^{3}}{\cos ^{4} \alpha \cdot \sin \alpha \cos \alpha}= \\
& =\frac{\pi r^{3}\left(\cos ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\right)^{3}}{3 \cos ^{2} \alpha \sin \alpha\left(\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}\right)^{3}}=\frac{\pi r^{3}\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)^{6}}{3 \cos ^{2} \alpha \sin \alpha\left(\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}\right)^{3}}=
\end{aligned}
$$

$$
=\frac{\pi r^{3}\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)^{3}}{3 \cos ^{2} \alpha \sin \alpha\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)^{3}}=\frac{\pi r^{3}\left(1+\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\right)^{3}}{3 \cos ^{2} \alpha \sin \alpha\left(1-\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\right)^{3}}=
$$

$$
=\frac{\pi r^{3}\left(1+\operatorname{tg} \frac{\alpha}{2} \cdot \operatorname{tg} \frac{\pi}{4}\right)^{3}}{3 \cos ^{2} \alpha \sin \alpha\left(\operatorname{tg} \frac{\pi}{4}-\operatorname{tg} \frac{\alpha}{2}\right)^{3}}=\frac{\pi r^{3}}{3 \cos ^{2} \alpha \sin \alpha} \cdot \frac{1}{\operatorname{tg}^{3}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}=\frac{\pi r^{3} \operatorname{ctg}^{3}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}{3 \cos ^{2} \alpha \sin \alpha} .
$$

$$
\text { Answer: } \frac{\pi r^{3} \operatorname{ctg}^{3}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}{3 \cos ^{2} \alpha \sin \alpha}
$$