A) Equilateral
B) Isosceles
C) Right angled
D) None of these
Correct Answer: B
Solution :
\[\cos A=\frac{\sin B}{2\sin C}\Rightarrow \frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=\frac{b}{2c}\] Þ \[{{b}^{2}}+{{c}^{2}}-{{a}^{2}}-{{b}^{2}}=0\]Þ\[{{c}^{2}}={{a}^{2}}\].
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