A) 9 : 7
B) 7 : 1
C) 5 : 3
D) 3 : 1
Correct Answer: B
Solution :
\[\frac{a}{b}=\frac{\sin \,A}{\sin \,B}\] \[\Rightarrow \,\,\,\frac{\sqrt{3}+1}{\sqrt{3}-1}\,\,=\,\,\frac{\sin (120{}^\circ -\theta )}{\sin \,\theta }\,\] \[\Rightarrow \,\,\,\frac{\sqrt{3}+1}{\sqrt{3}-1}\,\,=\,\,\frac{\frac{\sqrt{3}}{2}\cos \,\theta +\,\frac{1}{2}\sin \,\theta )}{\sin \,\theta }\,\] \[\Rightarrow \,\,\,\frac{\sqrt{3}+1}{\sqrt{3}-1}\,\,-\,\,\frac{1}{2}=\frac{\sqrt{3}}{2}\,\cot \,\theta \] \[\Rightarrow \,\,\,\frac{3+2\sqrt{3}}{2}=\frac{\sqrt{3}}{2}\,\cot \,\theta \] \[\Rightarrow \,\,\,\cot \,\theta \,=\,\sqrt{3}+2\] \[\Rightarrow \,\,\,\tan \,\theta \,\,=\,\,\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=2-\sqrt{3}\] \[\theta \,\,=\,\,15{}^\circ \] \[\therefore \,\,\,\,\angle A=120{}^\circ -15{}^\circ =105{}^\circ \] \[\therefore \,\,\,\frac{\angle A}{\angle B}=\frac{105}{15}\,=\,\frac{7}{1}\,=\,7:1\]
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