Two cones have a common vertex, and the generatrix of the first cone is the height of the second. The angle at the vertex of the axial section of the first cone is $\arccos \frac{1}{3}$, and that of the second is $-120^{\circ}$. Find the angle between the generatrices at which the lateral surfaces of the cones intersect.
Since $120^{\circ}>\arccos \frac{1}{\frac{1}{3}}$, the height of the first cone is inside the second (Fig.1). Let $A$ be the common vertex of the cones, $O$ be the center of the base of the second cone, $A O=1, A M$ and $A N$ be the common generators of the cones. Through the line $M N$, we draw a plane perpendicular to the height of the first cone and intersecting it at point $Q$. Then $Q$ is the center of the base of a cone similar to the first cone relative to point $A$ and $A M=A N=\frac{A O}{\cos \angle O A M}=2$. From

$$
A Q=A M \cos \angle Q A M=2 \cos \left(\frac{1}{2} \arccos \frac{1}{3}\right)=\frac{2 \sqrt{2}}{\sqrt{3}}
$$

Let $P$ be the midpoint of segment $M N$. Then points $O, A, Q$ and $P$ lie in the same plane (Fig.2), and since $\angle A O P=\angle A Q P$ $=90^{\circ}$, these points lie on a circle with diameter $A P$. Therefore, $A P=\frac{O Q}{\sin \angle O A Q}$. Let $K$ be a point on the extension of segment $A O$ beyond point $O$, such that $O K=A O=1$. Then $Q O$ is the median of the right triangle $A Q K$, drawn from the vertex of the right angle. Thus, $O Q=\frac{1}{2} A K=1$. Therefore,

$$
A P=\frac{1}{\sin \angle O A Q}=\sqrt{3}
$$

Therefore, $\angle M A P=30^{\circ}$ and $\angle M A N=60^{\circ}$.

## Answer

$60^{\circ}$.