4. Four rays originate from a point $O$ in space, and the angles between each pair of rays are equal. Find this angle.


4. Solution 1 As shown in the figure, for the regular tetrahedron $ABCD$, the angles between the lines connecting the center $O$ to each vertex are equal, so $\angle AOD$ is the angle we are looking for.
If the edge length is $a$, then $OA = OD = \frac{\sqrt{6}}{4} a$,
$$
\cos \angle AOD = \frac{OA^2 + OD^2 - AD^2}{2 \cdot OA \cdot OD} = 1 - \frac{AD^2}{2 \cdot OA^2} = 1 - \frac{a^2}{2 \times \left(\frac{\sqrt{6}}{4} a\right)^2} = -\frac{1}{3}.
$$

Thus, the angle we are looking for is $\pi - \arccos \frac{1}{3}$.
Solution 2 As shown in the figure, for the center $O$ of a cube to the four vertices $A, B, C, D$, the angles between the lines connecting the center $O$ to each vertex are equal, so $\angle AOD$ is the angle we are looking for. If the edge length is $a$, then $OA = OD = \frac{\sqrt{3}}{2} a, AD = \sqrt{2} a$,
$$
\cos \angle AOD = \frac{OA^2 + OD^2 - AD^2}{2 \cdot OA \cdot OD} = 1 - \frac{AD^2}{2 \cdot OA^2} = 1 - \frac{2a^2}{2 \times \frac{3}{4} a^2} = -\frac{1}{3},
$$

Thus, the angle we are looking for is $\pi - \arccos \frac{1}{3}$.