A) \[60{}^\circ \]
B) \[120{}^\circ \]
C) \[30{}^\circ \]
D) \[45{}^\circ \]
E) \[90{}^\circ \]
Correct Answer: B
Solution :
\[\because \] \[(\sqrt{3}-1)a=2b\Rightarrow \frac{a}{b}=\frac{2}{\sqrt{3}-1}\] and \[A=3B\] We know that \[\frac{\sin A}{a}=\frac{\sin B}{b}\] \[\Rightarrow \] \[\frac{\sin 3B}{\sin B}=\frac{2}{\sqrt{3}-1}\] \[\Rightarrow \] \[\frac{3\sin B-4{{\sin }^{3}}B}{\sin B}=\frac{2}{\sqrt{3}-1}\] \[\Rightarrow \] \[3-4{{\sin }^{2}}B=\frac{2(\sqrt{3}+1)}{3-1}\] \[\Rightarrow \] \[3-4{{\sin }^{2}}B=\sqrt{3}+1\] \[\Rightarrow \] \[3-\sqrt{3}-1=4{{\sin }^{2}}B\] \[\Rightarrow \] \[\frac{2-\sqrt{3}}{4}={{\sin }^{2}}B\] \[\Rightarrow \] \[{{\left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right)}^{2}}={{\sin }^{2}}B\Rightarrow \sin B=\frac{\sqrt{3}-1}{2\sqrt{2}}\] \[\Rightarrow \] \[B=15{}^\circ \]and\[A=45{}^\circ \] \[\therefore \]\[C=180{}^\circ -A-B=180{}^\circ -60{}^\circ =120{}^\circ \]
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