12.285. The angle between the height and the slant height of a cone is $\alpha$. A plane passing through the vertex of the cone forms an angle $\beta$ with the height ( $\beta<\alpha$ ). In what ratio does this plane divide the circumference of the base?
## Solution.

Let SO be the height of the given cone, $\triangle A S B$ be its section by a plane (Fig. 12.148), $\angle O S B=\alpha$.

If $C$ is the midpoint of $A B$, then $S C \perp A B$, $\dot{O} C \perp A B, \angle B O C=\frac{1}{2} \angle A O B$.

Draw a perpendicular $O D$ from point $O$ to the plane $A S B$. Then $D C$ is the projection of $O C$ onto the plane $A S B$, $O C \perp A B \Rightarrow D C \perp A B$ and point $D$ lies on the line $S C$ and $S C$ is the projection of $S O$ onto the plane $A S B, \angle O S C=\beta$.

In $\triangle S O B\left(\angle S O B=\frac{\pi}{2}\right): S O=O B \operatorname{ctg} \alpha$.

In $\triangle S O C\left(\angle S O C=\frac{\pi}{2}\right): S O=O C \operatorname{ctg} \beta \Rightarrow O B \operatorname{ctg} \alpha=O C \operatorname{ctg} \beta ; \frac{O C}{O B}=\frac{\operatorname{tg} \beta}{\operatorname{tg} \alpha}$. Let $\angle B O C=\gamma$, then from

$$
\begin{aligned}
& \triangle O C B\left(\angle O C B=\frac{\pi}{2}\right): \cos \gamma=\frac{O C}{O B}=\frac{\operatorname{tg} \beta}{\operatorname{tg} \alpha} ; \gamma=\arccos \frac{\operatorname{tg} \beta}{\operatorname{tg} \alpha} \Rightarrow \cup A E B=2 \gamma= \\
& =2 \arccos \frac{\operatorname{tg} \beta}{\operatorname{tg} \alpha} ; \cup A F B=2 \pi-2 \arccos \frac{\operatorname{tg} \beta}{\operatorname{tg} \alpha} ; \frac{\cup A E B}{\cup A F B}=\frac{\arccos \frac{\operatorname{tg} \beta}{\operatorname{tg} \alpha}}{\pi-\arccos \frac{\operatorname{tg} \beta}{\operatorname{tg} \alpha}} \\
& \text { Answer: } \frac{\arccos \frac{\operatorname{tg} \beta}{\operatorname{tg} \alpha}}{\pi-\arccos \frac{\operatorname{tg} \beta}{\operatorname{tg} \alpha}}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1040.jpg?height=704&width=510&top_left_y=80&top_left_x=718)

Fig. 12.148