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Geometry Difficulty 4.6 AIME Find the answer

2. In △ABC\triangle A B C, it is known that ∠A=120∘,AB=5,BC\angle A=120^{\circ}, A B=5, B C =7. Then sin⁡Bsin⁡C=()\frac{\sin B}{\sin C}=(\quad).

Pick one

Solution

2. D.

By the Law of Sines, we have
sin⁡C=casin⁡A=57×32=5314. Therefore, cot⁡C=1153. Hence, sin⁡Bsin⁡C=sin⁡(A+C)sin⁡C=sin⁡A⋅cos⁡C+cos⁡A⋅sin⁡Csin⁡C=sin⁡A⋅cot⁡C+cos⁡A=32×1153−12=35. \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}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.