A) \[{{a}^{2}}\]
B) \[ac\]
C) \[bc\]
D) None of these
Correct Answer: A
Solution :
We have,\[\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ac}\] On multiplying both sides by\[abc\] \[\Rightarrow \]\[2bc\,\,\cos \,\,A+ac\,\,\cos \,\,B+2ab\,\,\cos \,\,C={{a}^{2}}+{{b}^{2}}\] \[\Rightarrow \]\[({{b}^{2}}+{{c}^{2}}-{{a}^{2}})+\frac{({{c}^{2}}+{{a}^{2}}-{{b}^{2}})}{2}\] \[+({{a}^{2}}+{{b}^{2}}-{{c}^{2}})={{a}^{2}}+{{b}^{2}}\] \[\Rightarrow \] \[({{c}^{2}}+{{a}^{2}}-{{b}^{2}})=2{{a}^{2}}-2{{b}^{2}}\] \[\Rightarrow \] \[{{b}^{2}}+{{c}^{2}}={{a}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
