150. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. A plane passing through $A$ 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 $N$. Determine the measure of the dihedral angle between the planes $A C_{1} K$ and $A C_{1} N$.
150. Let the plane $A K N$ touch the sphere at point $P$, and the line $4 P$ intersect $N K$ at point $M$ (Fig. 29). Then the plane $C_{1} N A$ is the bisector plane of the dihedral angle formed by the planes $D_{1} C_{1} A$ and $C_{1} M A$ (the planes $D_{1} A N$ and $A N M$ touch the sphere, and the planes $D_{1} C_{1} A$ and $C_{1} M A$ pass through its center).

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-090.jpg?height=377&width=369&top_left_y=523&top_left_x=137)

Fig. 29.

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-090.jpg?height=457&width=461&top_left_y=445&top_left_x=625)

Fig. 30.

Similarly, the plane $C_{1} K A$ is the bisector plane of the dihedral angle formed by the planes $M C_{1} A$ and $C_{1} B_{1} A$. Thus, the dihedral angle between the planes $A C_{1} K$ and $A C_{1} N$ is half the dihedral angle between the planes $A D_{1} C_{1}$ and $A B_{1} C_{1}$, which is equal to $2 \pi / 3$.

Answer: $\pi / 3$