A) \[{{c}^{2}}-3c-7=0\]
B) \[{{c}^{2}}+3c+7=0\]
C) \[{{c}^{2}}-3c+7=0\]
D) \[{{c}^{2}}+3c-7=0\]
Correct Answer: A
Solution :
\[\cos A=\frac{({{b}^{2}}+{{c}^{2}}-{{a}^{2}})}{2bc}\] \[\Rightarrow \cos {{60}^{o}}=\frac{1}{2}=\frac{9+{{c}^{2}}-16}{2.3c}\]Þ \[{{c}^{2}}-3c-7=0\].
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