A) \[\sqrt{3}-1\]
B) \[\sqrt{3}\]
C) \[\sqrt{3}+1\]
D) None of these
Correct Answer: C
Solution :
We have \[\frac{b}{\sin B}=\frac{c}{\sin C}\]Þ \[\sin B=\frac{b\sin C}{c}\] Þ \[\sin B=\frac{2\sin {{60}^{o}}}{\sqrt{6}}=\frac{2}{\sqrt{6}}.\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}}\]Þ\[B={{45}^{o}}\],\[(\because B\ne {{135}^{o}})\] \ \[A={{180}^{o}}-(B+C)={{75}^{o}}\] Now\[\frac{\sin A}{a}=\frac{\sin B}{b}\]Þ\[a=\frac{b\sin A}{\sin B}=\frac{2\sin {{75}^{o}}}{\sin {{45}^{o}}}=\sqrt{3}+1\].
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