12.246. Find the angle between the generatrix of the cone and the plane of the base, if the lateral surface of the cone is equal to the sum of the areas of the base and the axial section.
## Solution.

Let $R$ be the radius of the base, $l$ the slant height, $H$ the height of the given cone, and $\alpha$ the angle we are looking for between the slant height and the base plane.

Then $H=R \operatorname{tg} \alpha, l=\frac{R}{\cos \alpha}$.

The lateral surface area of the cone $S_{1}=\pi R l=\frac{\pi R^{2}}{\cos \alpha}$. The base area $S_{2}=\pi R^{2}$, and the area of the axial section $S_{3}=R H=R^{2} \operatorname{tg} \alpha$.

Since $S_{1}=S_{2}+S_{3}$, we have $\frac{\pi R^{2}}{\cos \alpha}=\pi R^{2}+R^{2} \operatorname{tg} \alpha$:

$$
\pi=\pi \cos \alpha+\sin \alpha ; \frac{\sin \alpha}{1-\cos \alpha}=\pi ; \operatorname{ctg} \frac{\alpha}{2}=\pi ; \alpha=2 \operatorname{arcctg} \pi
$$

Answer: $2 \operatorname{arcctg} \pi$.