Question

In $△ABC$, if $\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ then

• A. $A={60}^o$
• B. $B={60}^o$
• C. $C={60}^o$
• D. $A={90}^o$

Answer 1

The correct option is A $A={60}^o$

$\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$

$\frac{2a+b+c}{a^2+ab+ac+bc}=\frac{3}{a+b+c}$

$(2a+b+c)(a+b+c)=3(a^2+ab+ac+bc)$

$2a^2+2ab+2ac+ab+b^2+bc+ac+bc+c^2$

$=3a^2+3ab+3ac+3bc$

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

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

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

$cosA=\frac{1}{2}$

Also, $cos{60}^{\circ }=\frac{1}{2}$

on comparing angles

$A={60}^{\circ }$