4. On the base plane of a cone with height $h$ and base radius $R$, a point (outside the cone) is given, which is at a distance of $aR$ from the circumference of the base. Find the angle between the tangent planes to the cone passing through the given point.

First method:

Let $L$ be the given point, $B$ and $C$ be the points of tangency of the planes with the circumference of the base, $\varphi=\angle B A O, \psi=\angle S A B, \alpha=\angle S A O$. The point $K$ is the intersection of the chord $B C$ and the line $A O$, where $O$ is the center of the base of the cone. Then $\sin \varphi=\frac{B O}{O A}=\frac{1}{a+1}$,

$B A=R \sqrt{a^{2}+2 a}, B K=R \cos \varphi=\frac{R \sqrt{a^{2}+2 a}}{a+1}, K A=B A \cos \varphi=\frac{R\left(a^{2}+2 a\right)}{a+1}$.
Let $S$ be the vertex of the cone. From the right triangle $\triangle S O A$ we find $\operatorname{tg} \alpha=$ $\frac{h}{(a+1) R}$. Therefore, $\sin \alpha=\frac{h}{\sqrt{h^{2}+(a+1)^{2} R^{2}}}$. Let $D$ be the point of the base of the perpendiculars dropped from points $B$ and $C$ to the line $S A$. Then $\delta=\angle B D C$ is the desired angle. From $\triangle D K A$ we find

![](https://cdn.mathpix.com/cropped/2024_05_06_778d240aea7dd323243eg-5.jpg?height=396&width=575&top_left_y=324&top_left_x=1939)

$$
K D=K A \sin \alpha=\frac{R h\left(a^{2}+2 a\right)}{(a+1) \sqrt{h^{2}+(a+1)^{2} R^{2}}}
$$

$$
\operatorname{tg}(\delta / 2)=\operatorname{tg} \angle K D B=\frac{B K}{K D}=\frac{\sqrt{h^{2}+(a+1)^{2} R^{2}}}{h \sqrt{a^{2}+2 a}}
$$

from which $\delta=2 \operatorname{arctg} \frac{\sqrt{h^{2}+(a+1)^{2} R^{2}}}{h \sqrt{a^{2}+2 a}}$.

Second method: By the cosine theorem for the trihedral angle $S A B C$ we obtain ${ }^{1}$ :

$$
\cos \delta=\frac{\cos 2 \varphi-\cos ^{2} \psi}{\sin ^{2} \psi}
$$

where $\cos \varphi=\frac{\sqrt{a^{2}+2 a}}{a+1}, \sin \psi=\frac{S B}{A S}=\frac{\sqrt{h^{2}+R^{2}}}{h^{2}+R^{2}(a+1)^{2}}$.

Since $h=R, a=2$, we obtain the answer.
Answer: $2 \operatorname{arctg}(\sqrt{5} / 2)$.

Answer to the variant: $5-2: 2 \operatorname{arctg}(\sqrt{2} / \sqrt{3})$, (here $h=2 R, a=1)$).