11.5. In a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, two diagonals of the lateral faces $A_{1} C_{1}$ and $C_{1} D$ are drawn. Find the angle between them, given that the diagonal $A_{1} C_{1}$ is equal in length to one of the edges of the parallelepiped, and the diagonal $C_{1} D$ forms an angle of 30 degrees with this same edge.
Answer: $\arccos \frac{\sqrt{3}}{6}$.

Solution. It is clear that the diagonal $A_{1} C_{1}$ can only be equal to the edges $A A_{1}=B B_{1}=$ $C C_{1}=D D_{1}$, because otherwise the leg of the right triangle $A_{1} D_{1} C_{1}$ would be equal to its hypotenuse. Let $D_{1} C_{1}=a$. Then, in the triangle $D D_{1} C_{1}$, by the condition $\angle D_{1} D C_{1}=30^{\circ}$, and $D C_{1}=2 a$. Therefore, $C_{1} A_{1}=D D_{1}=a \sqrt{3}$. Hence, $A_{1} D_{1}=a \sqrt{2}$ (from the right triangle $A_{1} D_{1} C_{1}$) and $A_{1} D=a \sqrt{5}$ (from the right triangle $A_{1} D_{1} D$). Now we find the cosine of the angle $A_{1} C_{1} D$ using the cosine theorem.

## Criteria.

7 points. Complete and justified solution.

Minus 2 points. The choice of the lateral edge of the parallelepiped is not justified.

Minus 1 point. For one computational error that does not affect the logic of the solution.

$\mathbf{0}$ points. In all other cases.