**Zaslavsky A.**.

A sphere is circumscribed around a regular tetrahedron $ABCD$. On its faces, as bases, regular pyramids $ABCD', ABD'C, ACD'B, BCD'A$ are constructed outwardly, with their vertices lying on this sphere. Find the angle between the planes $ABC_1$ and $ACD'$.
Consider a cube, vertices $A, B, C$ and $D$ of which coincide with the vertices of a tetrahedron (see figure). Since the sphere circumscribed around the cube is also circumscribed around the tetrahedron, $A^{\prime}, B^{\prime}, C^{\prime}$ and $D^{\prime}$ are the remaining four vertices of this cube, and the specified planes are the faces of the cube.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-01.jpg?height=283&width=349&top_left_y=870&top_left_x=846)

## Answer

$90^{\circ}$.

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Problem