3. A right circular cone is given with vertex $V$, the center of the base $S$, and the angle at the vertex of the axial section $\beta$. Two tangent planes to the cone touch the cone along the generators $V A$ and $V B$, where points $A$ and $B$ lie on the base. If the angle between these planes is $\alpha$, determine $\measuredangle A S B$.
Solution. The given tangent planes to the cone intersect the plane of the base along lines that are tangents to the base circle at points $A$ and $B$. Let $K$ be the intersection of these tangents. Triangles $A K V$ and $B K V$ have right angles at $A$ and $B$, respectively, a common hypotenuse $K V$, and equal legs $A K$ and $B K$, so they are congruent. Let $L$ be the common foot of the altitude corresponding to the hypotenuse of these triangles.

Then $\measuredangle A L B=\alpha$ and $\measuredangle A V S=\frac{\beta}{2}$. Let $\measuredangle A S B=\varphi$, the radius of the base be $r$, and the midpoint of segment $A B$ be $M$ (see diagram).

![](https://cdn.mathpix.com/cropped/2024_06_05_d0eb44e15bc267091b4bg-07.jpg?height=598&width=454&top_left_y=290&top_left_x=1011)
Using the right triangles $A S M, A M L, A S V, A K S$, we get:

$$
A M=r \sin \frac{\varphi}{2}, A L=\frac{r \sin \frac{\varphi}{2}}{\sin \frac{\alpha}{2}}, A V=\frac{r}{\sin \frac{\beta}{2}}, A K=r \operatorname{tg} \frac{\varphi}{2}
$$

In the right triangle $A K V$, the following holds:

$$
A V \cdot A K=A L \sqrt{A K^{2}+A V^{2}}
$$

Substituting the previous expressions, we get an equation for determining the unknown angle $\varphi$, from which, after simplification, we obtain

$$
\sin \frac{\varphi}{2}=\frac{\cos \frac{\alpha}{2}}{\cos \frac{\beta}{2}}
$$