Two equal cones have a common vertex and touch along a common generatrix. The angle in the axial section of each cone is $60^{\circ}$. Find the angle between two planes, each of which touches the cones but does not pass through the common generatrix.
Let the planes of the bases of the cones intersect the perpendicular to the common tangent plane, passing through the common generatrix $P K$ of the cones with a common vertex $K$, at points $Q$ and $Q_1$ (Fig.1). From point $Q$, draw tangents to the circle of the base of the first cone. Let these tangents intersect the common tangent plane at points $A$ and $B$. Then the tangents to the circle of the base of the second cone, drawn from point $Q_1$, intersect the common tangent plane at the same points. Consider two triangular pyramids $A B P Q$ and $A B P Q_1$ with a common base $A B P$ (the "frames" of the cones). Note that the desired angle is the dihedral angle between the planes $A Q Q_1$ and $B Q Q_1$, and $A P B$ is its linear angle. Let $O$ be the center of the base of the first cone, $r$ be the radius of the base of the cone, and $M$ be the point of tangency of the circle of the base with the segment $A Q$ (Fig.2). Then

$$
K P = 2 O K = 2 r, \quad P Q = K P \sqrt{3} = 2 r \sqrt{3}.
$$

In the right triangle $A P Q$ (Fig.3), it is known that $P M = P K = 2 r, \quad Q P = 2 r \sqrt{3}$. Let $\angle A Q P = \phi$. Then

$$
\sin \phi = \frac{P M}{P Q} = \frac{1}{\sqrt{3}}, \quad \cos \phi = \frac{\sqrt{2}}{\sqrt{3}}, \quad \tan \phi = \frac{1}{\sqrt{2}}, \quad A P = P Q \tan \phi = 2 r \sqrt{3} \cdot \frac{1}{\sqrt{2}} = r \sqrt{6}.
$$

If $\gamma$ is the desired angle, then

$$
\cos \frac{\gamma}{2} = \frac{K P^p}{A P^p} = \frac{2 r}{4 \sqrt{6}} = \sqrt{\frac{2}{3}}
$$

Then $\sin \frac{\gamma}{2} = \frac{1}{\sqrt{3}}$. Therefore,

$$
\cos \gamma = \cos 2 \frac{\gamma}{2} - \sin 2 \frac{\gamma}{2} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}
$$

## Answer

$2 \operatorname{arctg} \frac{\sqrt{2}}{2} = 2 \arccos \sqrt{\frac{2}{3}} = \arccos \frac{1}{3}$.