Problem 1205

1. Draw the regular pentagon and label the vertices: Let $ABCDE$ be a regular pentagon with vertices labeled in a clockwise or counterclockwise manner.

2. Identify the intersection point: Let $S$ be the intersection point of the diagonals $AC$ and $BD$.

3. Determine the internal angles of the pentagon: In a regular pentagon, each internal angle is given by:
$$\text{Internal angle} = \frac{(n-2) \times 180^\circ}{n} = \frac{(5-2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = 108^\circ$$

4. Identify the angles at the intersection: Since $ABCDE$ is a regular pentagon, the diagonals $AC$ and $BD$ intersect at point $S$. We need to find the acute angle at this intersection, specifically $\angle CSD$.

5. Use symmetry and properties of the regular pentagon: In a regular pentagon, the diagonals intersect in such a way that they divide the internal angles into equal parts. Therefore, $\angle SCD$ and $\angle CDS$ are each half of the internal angle at $C$ and $D$ respectively:
$$\angle SCD = \angle CDS = \frac{108^\circ}{2} = 54^\circ$$

6. Calculate the remaining angle: The sum of the angles in triangle $SCD$ must be $180^\circ$. Therefore, the remaining angle $\angle CSD$ can be calculated as:
$$\angle CSD = 180^\circ - \angle SCD - \angle CDS = 180^\circ - 54^\circ - 54^\circ = 72^\circ$$

7. Determine the acute angle: Since $\angle CSD = 72^\circ$ is the angle at the intersection of the diagonals, and it is less than $90^\circ$, it is indeed the acute angle we are looking for.

$$\boxed{72^\circ}$$