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\[\angle A={{90}^{o}}\].