Question

If in a triangle $ABC,\frac{2cosA}{a}+\frac{cosB}{b}+\frac{2cosC}{c}=\frac{a}{bc}+\frac{b}{ac}.Find\angle A.$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is A $\frac{\pi }{2}$

$\frac{2\cos A}{a}+\frac{2\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ac}$

using $\cos A=\frac{b^2+c^2-a^2}{2bc},....$ (cosine rule)

$\Rightarrow \frac{2}{a}\left(\frac{b^2+c^2-a^2}{2bc}\right)+\frac{1}{b}\left(\frac{c^2+a^2-b^2}{2ca}\right)+\frac{2}{c}\left(\frac{a^2+b^2-c^2}{2ab}\right)=\frac{a}{bc}+\frac{b}{ac}$

$\Rightarrow \frac{2(b^2+c^2-a^2)}{2abc}+\frac{c^2+a^2-b^2}{2abc}+\frac{2(a^2+b^2-c^2)}{2abc}=\frac{a}{bc}+\frac{b}{ac}$

Taking LCM

$\frac{2(b^2+c^2-a^2)+c^2+a^2-b^2+2(a^2+b^2-c^2)}{2abc}=\frac{a^2+b^2}{abc}$

$\Rightarrow 2b^2+2c^2-2a^2+c^2+a^2-b^2+2a^2+2b^2-2c^2=2a^2+2b^2$

$\Rightarrow -2a^2+c^2+a^2-b^2+2b^2=0\Rightarrow -a^2+c^2+b^2=0\Rightarrow b^2+c^2=a^2$

Above expression is pythagoras theorem

$\Rightarrow$ $a$ is hypotenuse of triangle $ABC$

$\Rightarrow$ opposite angle is ${90}^o$

$\Rightarrow$ $\angle A={90}^o=\frac{\pi }{2}$