3.102. The ratio of the volume of a sphere inscribed in a cone to the volume of a circumscribed sphere is $k$. Find the angle between the slant height of the cone and the plane of its base and the permissible values of $k$.
3.102. Let's depict the axial section of the figure; $S A=S B$ are the generators of the cone, $S O$ is the height of the cone, $O_{1}$ and $O_{2}$ are the centers of the circumscribed and inscribed spheres (Fig. 3.102), the ratio of their volumes is $k$; we need to find $\angle S A O$. Let $R$ and $r$ be the radii of the spheres, $\angle S A O=\alpha, A O=x$; then $\angle O O_{1} B=\angle A S B=180^{\circ}-2 \alpha$, $\angle O_{2} A O=\frac{\alpha}{2}$. From $\triangle O_{1} O B$ and $\triangle O_{2} O A$ we find

$$
R=\frac{x}{\sin \left(180^{\circ}-2 \alpha\right)}=\frac{x}{\sin 2 \alpha}, r=x \operatorname{tg} \frac{\alpha}{2}
$$

Thus,

$$
\frac{r}{R}=\frac{x \operatorname{tg} \frac{\alpha}{2} \sin 2 \alpha}{x}=\frac{(1-\cos \alpha) \cdot 2 \sin \alpha \cos \alpha}{\sin \alpha}=2\left(\cos \alpha-\cos ^{2} \alpha\right)
$$

Since $\quad \frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R^{3}}=k$, then $\frac{r}{R}=\sqrt[3]{k} . \quad$ Solving the equation

$2 \cos \alpha-2 \cos ^{2} \alpha=\sqrt[3]{k}$, we get

$$
\cos \alpha=\frac{1 \pm \sqrt{1-2 \sqrt[3]{k}}}{2}, \text { where } 1-2 \sqrt[3]{k} \geq 0, \text { hence } 0 \leq k \leq \frac{1}{8}
$$

Therefore, $\angle S A O=\alpha=\arccos \frac{1 \pm \sqrt{1-2 \sqrt[3]{k}}}{2}, 0<k \leq \frac{1}{8}$.