$32 \cdot 58$ In an acute triangle $\triangle A B C$, the measures of the three interior angles are all prime numbers. Then such a triangle
(A) has only one, and it is an isosceles triangle.
(B) has at least two, and all are isosceles triangles.
(C) has only one, but it is not an isosceles triangle.
(D) has at least one, and among them, there are non-isosceles triangles.
(2nd China Tianjin "Chinese Youth Cup" Mathematics Invitational Competition, 1987)
[Solution] Let the three interior angles of $\triangle ABC$ be $x, y, z$, thus
$$
x+y+z=180,
$$

Furthermore, $x, y, z$ are all prime numbers. Except for the prime number 2, all other prime numbers are odd. The sum of three odd numbers being even is impossible.
Therefore, one of $x, y, z$ must be 2. Let's assume it is $x$.
Thus, $y+z=178$, and $y, z$ are prime numbers less than 90. Hence, only $y=z=89$ satisfies the condition.
If one of $y, z$ is less than 89, then the other must be at least 90, which contradicts the problem's conditions.

Therefore, only the set $(2,89,89)$ fits.
Hence, the answer is $(A)$.