2. The three altitudes of a triangle are $2.4, 3, 4$. This triangle is ( ).
(A) acute triangle
(1)）right triangle
(C) obtuse triangle
(i) cannot be determined
2. (B).

Let the area of this triangle be $S$. Then its three sides are $\frac{2 S}{2.4}$, $\frac{2 S}{3}$, $\frac{2 S}{4}$, where $\frac{2 S}{2.4}$ is the longest side. Let the largest angle opposite to this side be $A$. By the cosine rule, we get
$$
\begin{array}{l}
\cos A=\frac{\left(\frac{2 S}{3}\right)^{2}+\left(\frac{2 S}{4}\right)^{2}-\left(\frac{2 S}{2.4}\right)^{2}}{2\left(\frac{2 S}{3}\right)\left(\frac{2 S}{4}\right)}=0, \\
\therefore \angle A=90^{\circ} .
\end{array}
$$