12.261. A plane is drawn through two generators of a cone, the angle between which is $\alpha$. The area of the section is to the total surface area of the cone as $2: \pi$. Find the angle between the generator and the height of the cone.
## Solution.

Let $B O$ be the height of the cone, $\triangle A B C$ be the section of the cone, $\angle A B C=\alpha$, (Fig. 12.127), $A B=d$.

Then the area of the section is

$$
S_{1}=\frac{1}{2} A B^{2} \sin \angle A B C=\frac{1}{2} d^{2} \sin \alpha
$$

If the desired angle $A B C$ is $\beta$, then from

$\triangle A O B\left(\angle A O B=90^{\circ}\right): A O=d \sin \beta$.

The area of the total surface of the cone is

$$
\begin{aligned}
& S_{2}=\pi \cdot O A^{2}+\pi \cdot O A \cdot A B=\pi d^{2}(1+\sin \beta) \sin \beta \\
& \text { Then } \frac{S_{1}}{S_{2}}=\frac{\frac{1}{2} d^{2} \sin \alpha}{\pi d^{2}(1+\sin \beta) \sin \beta} \Rightarrow \frac{\sin \alpha}{2 \pi(1+\sin \beta) \sin \beta}=\frac{2}{\pi} \Leftrightarrow \\
& \Leftrightarrow 4 \sin ^{2} \beta+4 \sin \beta-\sin \alpha=0 \\
& \sin \beta=\frac{-2 \pm \sqrt{4+4 \sin \alpha}}{4}=\frac{-2 \pm 2 \sqrt{1+\cos \left(90^{\circ}-\alpha\right)}}{4}= \\
& =\frac{-1 \pm \sqrt{2 \cos ^{2}\left(45^{\circ}-\frac{\alpha}{2}\right)}}{2}=\frac{-1 \pm \sqrt{2} \cos \left(45^{\circ}-\frac{\alpha}{2}\right)}{2}
\end{aligned}
$$

Considering that the angle $\beta$ is acute, $\sin \beta>0 \Rightarrow$

$$
\begin{aligned}
& \sin \beta=\frac{\sqrt{2} \cos \left(45^{\circ}-\frac{\alpha}{2}\right)-1}{2}=\frac{\sqrt{2}}{2}\left(\cos \left(45^{\circ}-\frac{\alpha}{2}\right)-\frac{1}{\sqrt{2}}\right)= \\
& =\frac{\sqrt{2}}{2}\left(\cos \left(45^{\circ}-\frac{\alpha}{2}\right)-\cos 45^{\circ}\right)=\sqrt{2} \sin \frac{\alpha}{4} \sin \left(45^{\circ}-\frac{\alpha}{4}\right) \\
& \beta=\arcsin \left(\sqrt{2} \sin \frac{\alpha}{4} \sin \left(45^{\circ}-\frac{\alpha}{4}\right)\right) \\
& \text { Answer: } \arcsin \left(\sqrt{2} \sin \frac{\alpha}{4} \sin \left(45^{\circ}-\frac{\alpha}{4}\right)\right)
\end{aligned}
$$