12.103. A plane is drawn through two generators of a cone, the angle between which is $\alpha$. Find the ratio of the area of the section to the total surface area of the cone, if the generator of the cone makes an angle $\beta$ with the base plane.
## Solution.

The area of the section is $S_{\mathrm{ce4}}=\frac{1}{2} S A^{2} \sin \alpha$ (Fig. 12.103), the total surface area of the cone is $S_{\text {pol }}=S_{\text {bok }}+S_{\text {osn }}=\pi R l+\pi R^{2}=\pi A O \cdot S A+\pi A O^{2}$. From $\triangle S O A\left(\angle S O A=90^{\circ}\right) A O=S A \cos \beta$. Substituting:

$S_{\text {pol }}=\pi S A^{2}\left(\cos \beta+\cos ^{2} \beta\right)$. Then we finally get:

$$
\frac{S_{\mathrm{cey}}}{S_{\text {pol }}}=\frac{S A^{2} \sin \alpha}{2 \pi S A^{2}\left(\cos \beta+\cos ^{2} \beta\right)}=\frac{\sin \alpha}{4 \pi \cos \beta \cos ^{2} \frac{\beta}{2}}
$$

Answer: $\frac{\sin \alpha}{4 \pi \cos \beta \cos ^{2} \frac{\beta}{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-791.jpg?height=418&width=460&top_left_y=291&top_left_x=154)

Fig. 12.103

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-791.jpg?height=624&width=530&top_left_y=92&top_left_x=687)

Fig. 12.104