12.336. The slant height of the cone is equal to $l$ and forms an angle $\alpha$ with the height. A plane is drawn through two slant heights of the cone, the angle between which is $\beta$. Find the distance from this plane to the center of the sphere inscribed in the cone.
Solution.

Let $SO$ be the height of the given cone, $\triangle ASB$ be the section of the cone, $\angle ASO = \alpha, AS = l, \angle ASB = \beta$, the center $O_1$ of the inscribed sphere belongs to the segment $SO$ (Fig. 12.197).

$C$ is the midpoint of $AB$. Then $OC \perp AB$, $SC \perp AB$. Therefore, the line $AB$ is perpendicular to the plane $SCO$ and the plane

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1094.jpg?height=620&width=521&top_left_y=966&top_left_x=739)

Fig. 12.197 $SCO$ and $ASB$ are perpendicular, $SC$ is the line of their intersection.

In the plane $SCO$ - draw the perpendicular $O_1D$ to $SC$. Then the segment $O_1D$ is perpendicular to the plane $ASB$ and its length is the desired distance.

In $\triangle SOA\left(\angle SOA = \frac{\pi}{2}\right): AO = l \sin \alpha; SO = l \cos \alpha$.

In $\triangle SCA\left(\angle SCA = \frac{\pi}{2}\right): AC = l \sin \frac{\beta}{2}; SC = l \cos \frac{\beta}{2}$.

In $\triangle ACO\left(\angle ACO = \frac{\pi}{2}\right)$:

$CO = \sqrt{AO^2 - AC^2} = \sqrt{l^2 \sin^2 \alpha - l^2 \sin^2 \frac{\beta}{2}} = l \sqrt{\frac{1 - \cos 2\alpha}{2} - \frac{1 - \cos \beta}{2}} = $

$= l \sqrt{\frac{\cos \beta - \cos 2\alpha}{2}} = l \sqrt{\sin \left(\alpha + \frac{\beta}{2}\right) \sin \left(\alpha - \frac{\beta}{2}\right)}$.

Since the sphere inscribed in the cone touches the sides of the angle $SAO$, $AO_1$ is the bisector of this angle, $\angle OAO_1 = \frac{1}{2} \angle SAO = \frac{1}{2}\left(\frac{\pi}{2} - \alpha\right) = \frac{\pi}{4} - \frac{\alpha}{2}$.

In $\triangle AOO_1\left(\angle AOO_1 = \frac{\pi}{2}\right): OO_1 = AO \tan \angle OAO_1 = l \sin \alpha \tan\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$

$SO_1 = SO - OO_1 = l \cos \alpha - l \sin \alpha \tan\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = $

$= l \cdot \frac{\cos \alpha \cos \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) - \sin \alpha \sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)}{\cos \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)} = l \cdot \frac{\cos \left(\frac{\pi}{4} + \frac{\alpha}{2}\right)}{\sin \left(\frac{\pi}{4} + \frac{\alpha}{2}\right)} = l \cot\left(\frac{\pi}{4} + \frac{\alpha}{2}\right)$

$\Delta SDO_1 \sim \triangle SOC$ - by two angles.

Then $\frac{SC}{SO_1} = \frac{OC}{DO_1}$

$O_1D = \frac{SO_1 \cdot OC}{SC} = \frac{l \cot\left(\frac{\pi}{4} + \frac{\alpha}{2}\right) \cdot l \sqrt{\sin \left(\alpha + \frac{\beta}{2}\right) \sin \left(\alpha - \frac{\beta}{2}\right)}}{l \cos \frac{\beta}{2}}$

$= \frac{l \cot\left(\frac{\pi}{4} + \frac{\alpha}{2}\right) \sqrt{\sin \left(\alpha + \frac{\beta}{2}\right) \sin \left(\alpha - \frac{\beta}{2}\right)}}{\cos \frac{\beta}{2}}$

Answer: $\frac{l \cot\left(\frac{\pi}{4} + \frac{\alpha}{2}\right) \sqrt{\sin \left(\alpha + \frac{\beta}{2}\right) \sin \left(\alpha - \frac{\beta}{2}\right)}}{\cos \frac{\beta}{2}}$.