A) \[{{45}^{o}}\]
B) \[{{30}^{o}}\]
C) \[{{90}^{o}}\]
D) \[{{60}^{o}}\]
Correct Answer: C
Solution :
\[\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ca}\] Þ \[\frac{2({{b}^{2}}+{{c}^{2}}-{{a}^{2}})}{2abc}+\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2abc}+\frac{2({{a}^{2}}+{{b}^{2}}-{{c}^{2}})}{2abc}\]\[=\frac{a}{bc}+\frac{b}{ca}\] Þ\[\frac{3{{b}^{2}}+{{c}^{2}}+{{a}^{2}}}{2abc}\]\[=\frac{a}{bc}+\frac{b}{ca}\]Þ\[\frac{3b}{2ac}+\frac{c}{2ab}+\frac{a}{2bc}=\frac{a}{bc}+\frac{b}{ca}\] Þ \[{{b}^{2}}+{{c}^{2}}={{a}^{2}}\]. Hence\[\angle A={{90}^{o}}\].
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