12.267. The ratio of the total surface area of a cone to the area of its axial section is $k$. Find the angle between the height and the slant height of the cone and the permissible values of $k$.
## Solution.

Let $\alpha$ be the angle between the height and the slant height of the cone, $l$ be the slant height, and $r$ be the radius of the base of the cone.

Then $r = l \sin \alpha$ and the area of the total surface of the cone is

$$
S_{1} = \pi r^{2} + \pi r l = \pi l^{2} \sin ^{2} \alpha + \pi l^{2} \sin \alpha = \pi l^{2} \sin \alpha (\sin \alpha + 1)
$$

The angle at the vertex of the axial section is $2 \alpha$ and the area of the section is $S_{2} = \frac{1}{2} l^{2} \sin 2 \alpha$.

Thus,

$$
\begin{aligned}
& k = \frac{S_{1}}{S_{2}} = \frac{\pi l^{2} \sin \alpha (\sin \alpha + 1)}{\frac{1}{2} l^{2} \sin 2 \alpha} = \frac{\pi (\sin \alpha + 1)}{\cos \alpha} \Rightarrow \frac{1 + \sin \alpha}{\cos \alpha} = \frac{k}{\pi} \Leftrightarrow \\
& \Leftrightarrow \frac{\cos \alpha}{1 + \sin \alpha} = \frac{\pi}{k} \Leftrightarrow \frac{\sin \left(\frac{\pi}{2} - \alpha\right)}{1 + \cos \left(\frac{\pi}{2} - \alpha\right)} = \frac{\pi}{k} \Leftrightarrow \operatorname{tg}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \frac{\pi}{k} \Rightarrow \frac{\alpha}{2} - \frac{\pi}{4} = \\
& = -\operatorname{arctg} \frac{\pi}{k} \Leftrightarrow \alpha = \frac{\pi}{2} - 2 \operatorname{arctg} \frac{\pi}{k}
\end{aligned}
$$

where $0 < \alpha < \pi$.

Answer: $\frac{\pi}{2} - 2 \operatorname{arctg} \frac{\pi}{k}, k > \pi$.