2. In $\triangle A B C$, it is known that $\angle A=120^{\circ}, A B=5, B C$ =7. Then $\frac{\sin B}{\sin C}=(\quad)$.
(A) $\frac{8}{5}$
(B) $\frac{5}{8}$
(C) $\frac{5}{3}$
(D) $\frac{3}{5}$
2. D.

By the Law of Sines, we have
$$
\begin{array}{l}
\sin C=\frac{c}{a} \sin A=\frac{5}{7} \times \frac{\sqrt{3}}{2}=\frac{5 \sqrt{3}}{14} . \\
\text { Therefore, } \cot C=\frac{11}{5 \sqrt{3}} . \\
\text { Hence, } \frac{\sin B}{\sin C}=\frac{\sin (A+C)}{\sin C} \\
=\frac{\sin A \cdot \cos C+\cos A \cdot \sin C}{\sin C} \\
=\sin A \cdot \cot C+\cos A \\
=\frac{\sqrt{3}}{2} \times \frac{11}{5 \sqrt{3}}-\frac{1}{2}=\frac{3}{5} .
\end{array}
$$