Question Number 79453 by Vishal Sharma last updated on 25/Jan/20
$$\mathrm{If}\:\mathrm{in}\:\mathrm{a}\:\mathrm{triangle}\:{ABC} \\ $$$$\mathrm{2}\:\frac{\mathrm{cos}\:{A}}{{a}}+\frac{\mathrm{cos}\:{B}}{{b}}+\mathrm{2}\frac{\mathrm{cos}\:{C}}{{c}}\:=\:\frac{{a}}{{bc}}\:+\:\frac{{b}}{{ca}} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle}\:{A}\:\mathrm{is} \\ $$
Answered by $@ty@m123 last updated on 25/Jan/20
$$\mathrm{2}\:\frac{\mathrm{cos}\:{A}}{{a}}+\frac{\mathrm{cos}\:{B}}{{b}}+\mathrm{2}\frac{\mathrm{cos}\:{C}}{{c}}\:=\:\frac{{a}}{{bc}}\:+\:\frac{{b}}{{ca}} \\ $$$$\Rightarrow\mathrm{2}\:\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{abc}}+\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{abc}}+\mathrm{2}\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{abc}}\:=\:\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{abc}} \\ $$$$\Rightarrow\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)=\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\Rightarrow\angle{A}=\mathrm{90}^{\mathrm{o}} \\ $$