2. In the tetrahedron $A-B C D$,
$$
\begin{array}{l}
\sin \angle B A D=\cos \angle B C D, \\
\sin \angle A B D=\cos \angle C B D, \\
\sin \angle A D B=\cos \angle C D B .
\end{array}
$$

Then $\triangle A B D$ is ( ).
(A) an obtuse triangle
(B) a right triangle
(C) an acute triangle
(D) cannot be determined, depends on the shape of the tetrahedron
2. A.

As shown in Figure 1, from the fact that $\angle B A D$, $\angle A B D$, and $\angle A D B$ are the interior angles of $\triangle A B D$, we have
$$
\begin{array}{l}
\cos \angle B C D \\
=\sin \angle B A D>0, \\
\cos \angle C B D \\
=\sin \angle A B D>0, \\
\cos \angle C D B=\sin \angle A D B>0 .
\end{array}
$$

Therefore, $\angle B C D$, $\angle C B D$, and $\angle C D B$ are all acute angles.
Next, we prove that the largest angle in $\triangle A B D$ must be an obtuse angle.
Assume without loss of generality that $\angle B A D$ is the largest angle in $\triangle A B D$, then $\angle A B D$ and $\angle A D B$ are both acute angles. Thus, $90^{\circ}-\angle A B D$ and $90^{\circ}-\angle A D B$ are also acute angles.
From equations (1) and (2), we have
$$
\angle C B D=90^{\circ}-\angle A B D, \angle C D B=90^{\circ}-\angle A D B \text {. }
$$

In $\triangle A B D$,
$$
\begin{aligned}
& \angle B A D=180^{\circ}-\angle A B D-\angle A D B \\
= & \left(90^{\circ}-\angle A B D\right)+\left(90^{\circ}-\angle A D B\right) \\
= & \angle C B D+\angle C D B=180^{\circ}-\angle B C D>90^{\circ} .
\end{aligned}
$$

Therefore, $\triangle A B D$ must be an obtuse triangle.