A) \[a=b\]
B) \[b=c\]
C) \[c=a\]
D) \[2a=bc\]
Correct Answer: C
Solution :
\[2\cos A=\frac{\sin B}{\sin C}\Rightarrow \frac{2({{c}^{2}}+{{b}^{2}}-{{a}^{2}})}{2bc}=\frac{b}{c}\] \[\Rightarrow \] \[{{c}^{2}}={{a}^{2}}\Rightarrow \] \[=\pi {{R}^{2}}=4\pi \].
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