A straight prism has a rhombus as its base, with an acute angle of $60^{\circ}$. What angle does a plane form with the base if the intersection of this plane with the prism's lateral surface is a square?
I. solution. The angle $CDA$ of the rhombus $ABCD$ is $60^{\circ}$, and let the length of the sides of the rhombus be $d$; then the length of the shorter diagonal is also $d$, since the triangle $ACD$ is equilateral. The length of the longer diagonal of the rhombus is $d \sqrt{3}$.

Two perpendicular lines projected orthogonally onto a plane are perpendicular to each other if and only if one of the original two lines is parallel to the plane. Since the diagonals of a rhombus (specifically a square) are perpendicular to each other, one of the diagonals of the intersecting square is parallel to the base of the prism.

Consider a plane parallel to the base of the prism. Rotate this plane around one of the diagonals of the rhombus until it cuts a square from the lateral surface of the prism. This means that the length of the diagonal of the square is equal to the length of one of the diagonals of the rhombus. In the second diagram, we have depicted the prism in the projection plane of the non-parallel diagonal of the square, where $\alpha$ is the angle between the two planes. The question is, around which diagonal of the rhombus did we rotate the plane. Suppose the diagonal of the square is parallel to the shorter diagonal of the rhombus, $d$. Then, from the right triangle $FJH$, $\cos \alpha=\frac{d}{d}=1$, which is a contradiction (2.a diagram).

Therefore, the diagonal of the square can only be parallel to the longer diagonal of the rhombus, and in this case,

$$
\cos \alpha=\frac{d}{d \sqrt{3}}=\frac{1}{\sqrt{3}}, \quad \text { hence } \quad \alpha \approx 54.7^{\circ} \quad(2 . b) \text { diagram) }
$$

II. solution. We know that the area $T$ of a plane figure and the area $T_{v}$ of its orthogonal projection onto another plane are related by the following equation: $T_{v}=T \cos \alpha$, where $\alpha$ is the angle between the two planes. The area of the projection is the area of the rhombus: $T_{v}=\frac{d^{2} \sqrt{3}}{2}$, and the area of the square is: $T=\left(\frac{d \sqrt{3}}{\sqrt{2}}\right)^{2}$,

$$
\cos \alpha=\frac{T_{v}}{T}=\frac{d^{2} \sqrt{3}}{3 d^{2}}=\frac{\sqrt{3}}{3}, \quad \text { hence } \quad \alpha \approx 54.7^{\circ}
$$

We must also realize that the length of the diagonal of the square cannot be $d$.