Question

If in a triangle ABC $\frac{2cosA}{a}+\frac{cosB}{b}+\frac{2cosC}{c}=\frac{a}{bc}+\frac{b}{ca}$ then

• A. $A={90}^o$
• B. $B={90}^o$
• C. $C={90}^o$
• D. none of these

Answer 1

The correct option is A $A={90}^o$

$\frac{2cosA}{a}+\frac{cosB}{b}+\frac{2cosC}{c}=\frac{a^2+b^2}{abc}$
$\frac{2(b^2+c^2-a^2)+c^2+a^2-b^2+2(a^2+b^2-c^2)}{2abc}=\frac{a^2+b^2}{abc}$
$3b^2+a^2+c^2=2(a^2+b^2)$

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

$2bc.cosA=0$

Now $a,b,c\ne 0$.

Hence

$cosA=0=cos{90}^0$
$\therefore A={90}^0$.