Question

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

• A. $30$
• B. $60$
• C. $90$
• D. $120$

Answer 1

Answer: (C)

$90$

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

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

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

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

$3b^2+a^2+c^2=2a^2+2b^2$

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

$\therefore \angle A=\frac{\pi }{2}$