12.245. The ratio of the volume of a cone to the volume of a sphere inscribed in it is $k$. Find the angle between the slant height and the base plane of the cone and the permissible values of $k$.
## Solution.

Let $\triangle A B C$ be the axial section of the given cone (Fig. 12.112), $D$ the center of its base, $O$ the center of the inscribed sphere, $\angle B A D$ the angle we need to find between the generatrix and the base plane of the cone, $\angle B A D=\alpha$, and the radius of the base of the cone $O A=R$.

In $\triangle A D B\left(\angle A D B=90^{\circ}\right):$

$B D=A D \operatorname{tg} \angle B A D=R \operatorname{tg} \alpha$,

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0994.jpg?height=619&width=511&top_left_y=417&top_left_x=768)

Fig. 12.112

$\angle O A D=\frac{1}{2} \angle B A D=\frac{\alpha}{2}$.

In $\triangle A D O\left(\angle A D O=90^{\circ}\right): D O=A D \operatorname{tg} \angle O A D=R \operatorname{tg} \frac{\alpha}{2}$.

The volume of the cone $V_{1}=\frac{1}{3} \pi \cdot O A^{2} \cdot O B=\frac{1}{3} \pi R^{3} \operatorname{tg} \alpha$,

the volume of the sphere $V_{2}=\frac{4}{3} \pi \cdot D O^{3}=\frac{4}{3} \pi R^{3} \operatorname{tg}^{3} \frac{\alpha}{2} \Rightarrow$

$$
\begin{aligned}
& \Rightarrow \frac{V_{1}}{V_{2}}=\frac{\operatorname{tg} \alpha}{4 \operatorname{tg}^{3} \frac{\alpha}{2}}=k ; \frac{2 \operatorname{tg} \frac{\alpha}{2}}{4 \operatorname{tg}^{3} \frac{\alpha}{2}\left(1-\operatorname{tg}^{2} \frac{\alpha}{2}\right)}=k \Rightarrow \\
& \Rightarrow 2 k \operatorname{tg}^{4} \frac{\alpha}{2}-2 k \operatorname{tg}^{2} \frac{\alpha}{2}+1=0 ; \operatorname{tg}^{2} \frac{\alpha}{2}=\frac{k \pm \sqrt{k^{2}-2 k}}{2 k} \Rightarrow \\
& \Rightarrow\left\{\begin{array}{l}
k^{2}-2 k \geq 0, \\
k>0
\end{array} \Leftrightarrow k \geq 2 .\right.
\end{aligned}
$$

From this, $\alpha=2 \operatorname{arctg} \sqrt{\frac{k \pm \sqrt{k^{2}-2 k}}{2 k}}$.

Answer: $2 \operatorname{arctg} \sqrt{\frac{k \pm \sqrt{k^{2}-2 k}}{2 k}} ; k \geq 2$.