4.8. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. A plane passing through vertex $A$ and tangent to the sphere inscribed in the cube intersects the edges $A_{1} B_{1}$ and $A_{1} D_{1}$ at points $K$ and $N$,

Find the measure of the angle between the planes $A C_{1} K$ and $A C_{1} N$.
4.8. Let us first prove the following auxiliary statement. Let two planes, intersecting along the line $A X$, touch a sphere with center $O$ at points $F$ and $G$. Then $A O X$ is the bisector plane of the dihedral angle formed by the planes $A O F$ and $A O G$. Indeed, the points $F$ and $G$ are symmetric relative to the plane $A O X$.

Let the plane $A K N$ touch the inscribed sphere of a cube at point $P$, and the line $A P$ intersect $N K$ at point $M$. Applying the proven statement to the tangent planes passing through the line $N A$, we obtain that $A C_{1} N$ is the bisector plane of the dihedral angle formed by the planes $A C_{1} D_{\mathrm{i}}$ and $A C_{1} M$. Similarly, $A C_{1} K$ is the bisector plane of the dihedral angle formed by the planes $A C_{1} M$ and $A C_{1} B_{1}$. Therefore, the angle between the planes $A C_{1} N$ and $A C_{1} K$ is half the dihedral angle formed by the half-planes $A C_{1} D_{1}$ and $A C_{1} B_{1}$. Considering the projection onto a plane perpendicular to $A C_{1}$, we obtain that this dihedral angle, formed by the half-planes $A C_{1} D_{1}$ and $A C_{1} B_{1}$, is equal to $120^{\circ}$.