2. Given in a convex quadrilateral $A B C D$, $A B=5 \sqrt{2}$, $B C=\sqrt{13}$, $C D=3 \sqrt{13}$, $A D=7$, $\angle B C D=$ $120^{\circ}$. Then the degree of $\angle B A D$ is $(\quad)$.
(A) $120^{\circ}$
(B) $105^{\circ}$
(C) $135^{\circ}$
(D) $150^{\circ}$
2.C.

As shown in Figure 5, draw $B E \perp$ $A D$ and $B F \perp C D$ at points $E$ and $F$ respectively. Connect $B D$.
Let $B E=x$,
$$
A E=y \text{.}
$$

Since $\angle B C D=120^{\circ}$, we have
$$
\angle B C F=180^{\circ}-\angle B C D=60^{\circ} \text{.}
$$

In the right triangle $\triangle B C F$, we have
$$
\begin{array}{l}
B F=\sqrt{13} \sin 60^{\circ}=\frac{\sqrt{39}}{2}, \\
C F=\sqrt{13} \cos 60^{\circ}=\frac{\sqrt{13}}{2},
\end{array}
$$

Thus, $D F=C F+C D=\frac{\sqrt{13}}{2}+3 \sqrt{13}=\frac{7 \sqrt{13}}{2}$. 
In the right triangle $\triangle B D F$, by the Pythagorean theorem, we get
$$
B D^{2}=B F^{2}+D F^{2}=169 \text{.}
$$

In the right triangles $\triangle A B E$ and $\triangle B D E$, by the Pythagorean theorem, we have
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=50, \\
x^{2}+(y+7)^{2}=169 .
\end{array}\right.
$$

Solving these equations, we get $x=5, y=5$, hence
$$
A E=B E \text{.}
$$

Therefore, $\angle E A B=45^{\circ}$.
So, $\angle B A D=180^{\circ}-\angle E A B=135^{\circ}$.