2. The three sides of a triangle are all integers, and its perimeter equals 10. Then this triangle is ( ).
(A) a right triangle
(B) an obtuse triangle
(C) a triangle with exactly two equal sides
(D) a triangle with exactly one interior angle of $60^{\circ}$
2. C.

Decompose 10 into the sum of three positive integers, we have
$$
\begin{array}{l}
10=1+1+8=1+2+7=1+3+6=1+4+5 \\
=2+2+6=2+3+5=2+4+4=3+3+4
\end{array}
$$

There are a total of eight cases. According to the "sum of the lengths of any two sides of a triangle is greater than the length of the third side," only the groups $(2,4,4)$ and $(3,3,4)$ can form triangles. Since the two base angles of an isosceles triangle are acute, in the isosceles triangle with sides $2,4,4$, the angle opposite the smallest side 2 is also acute. In the isosceles triangle with sides $3,3,4$, since $3^{2}+3^{2}>4^{2}$, the angle opposite the base is also acute. Therefore, the isosceles triangles with sides $2,4,4$ and $3,3,4$ are both acute triangles, eliminating options (A) and (B). Furthermore, since an isosceles triangle has exactly one angle of $60^{\circ}$ when it becomes an equilateral triangle, which contradicts the conditions of sides $(2,4,4)$ and $(3,3,4)$, option (D) is also eliminated. The triangles with sides $(2,4,4)$ and $(3,3,4)$ are isosceles triangles with exactly two equal sides.