A) \[{{c}^{2}}-5c-9=0\]
B) \[{{c}^{2}}-4c-9=0\]
C) \[{{c}^{2}}-10c+25=0\]
D) \[{{c}^{2}}-5c-41=0\]
E) \[{{c}^{2}}-4c-41=0\]
Correct Answer: B
Solution :
Given, \[\angle A=60{}^\circ ,\text{ }a=5,\text{ }b=4\] \[\therefore \] \[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] \[\Rightarrow \] \[\cos 60{}^\circ =\frac{1}{2}=\frac{16+{{c}^{2}}-25}{8c}\] \[\Rightarrow \] \[4x={{c}^{2}}-9\] \[\Rightarrow \] \[{{c}^{2}}-4c-9=0\]
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