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Question Number 79453 by Vishal Sharma last updated on 25/Jan/20

$$\mathrm{If}\mathrm{in}\mathrm{a}\mathrm{triangle}ABC$$$$2\frac{\cos A}{a}+\frac{\cos B}{b}+2\frac{\cos C}{c}=\frac{a}{bc}+\frac{b}{ca}$$$$\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{angle}A\mathrm{is}$$

Answered by $@ty@m123 last updated on 25/Jan/20

$$2\frac{\cos A}{a}+\frac{\cos B}{b}+2\frac{\cos C}{c}=\frac{a}{bc}+\frac{b}{ca}$$$$\Rightarrow 2\frac{b^2+c^2-a^2}{2abc}+\frac{c^2+a^2-b^2}{2abc}+2\frac{a^2+b^2-c^2}{2abc}=\frac{a^2+b^2}{abc}$$$$\Rightarrow 2(b^2+c^2-a^2)+c^2+a^2-b^2+2(a^2+b^2-c^2)=2(a^2+b^2)$$$$\Rightarrow a^2+3b^2+c^2=2a^2+2b^2$$$$\Rightarrow a^2=b^2+c^2$$$$\Rightarrow \mathrm{\angle }A={90}^{\mathrm{o}}$$

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