Question

In a triangle $ABC$, if $\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ca}$,

then which of the following is correct?

• A. $A={90}^o$
• B. $B={90}^o$
• C. $C={90}^o$
• D. $C={75}^o$

Answer 1

Answer: (A)

$A={90}^o$

Given that $\frac{2\cos A}{a}+\frac{\cos B}{b}+\frac{2\cos C}{c}=\frac{a}{bc}+\frac{b}{ca}$
Using cosine rule, we get
$\frac{b^2+c^2-a^2}{abc}+\frac{a^2+c^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{abc}=\frac{a^2+b^2}{abc}$
.
$\Rightarrow \frac{2b^2+2c^2-2a^2+a^2+c^2-b^2+2a^2+2b^2-2c^2}{2abc}=\frac{a^2+b^2}{abc}$
.
$\Rightarrow \frac{a^2+c^2+3b^2}{2abc}=\frac{a^2+b^2}{abc}$
.
$\Rightarrow a^2=b^2+c^2$
.
Since $a,b,c$ satisfy the Pythagoras theorem, it is a right-angled triangle, with the angle opposite to $a$, that is $\angle A$ as ${90}^o$