Find the dihedral angles of a regular tetrahedron.

#
Let $A B C D$ be a regular tetrahedron, $O$ be the center of the face $A B C$, and $M$ be the midpoint of $B C$. Since $O M \perp B C$ and $D M \perp B C$, $O M D$ is the linear angle of the dihedral angle formed by the faces $A B C$ and $D B C$. Since $A B C$ and $D B C$ are equal equilateral triangles, $A M$ and $D M$ are their heights, and $O$ is the center of triangle $A B C$, then

$$
O M=\frac{1}{3} A M=\frac{1}{3} \cdot \frac{\alpha \sqrt{3}}{2}=\frac{\alpha \sqrt{3}}{6}
$$

From the right triangle $O M D$, we find that

$$
\cos \angle O M D=\frac{O M}{D M}=\frac{\frac{a \sqrt{3}}{\frac{\sqrt{3}}{3}}}{2}=\frac{1}{3}
$$

Similarly for the other dihedral angles of the regular tetrahedron.

## Answer

$\arccos \frac{1}{3}$