Question

In $△ABC$, if $\frac{1}{a+b}+\frac{1}{a+c}=\frac{3}{a+b+c}$ then prove that $\angle A={60}^o$

Answer 1

We have,

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

$\text{Then}\text{prove}\text{that}\text{.}$

$\angle A={60}^0$

$\text{Now,}$

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

$\Rightarrow \frac{a+c+a+b}{(a+b)(a+c)}=\frac{3}{a+b+c}$

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

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

$\Rightarrow 2a^2+2ab+2ac+ab+b^2+bc+ac+bc+c^2=3a^2+3ac+3a+3ab+3bc$

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

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

$1=\frac{b^2+c^2-a^2}{bc}$

$\text{on}\text{divide both side by 2}\text{.}$

$\frac{1}{2}=\frac{b^2+c^2-a^2}{2bc}$

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

$CosA=Cos\frac{\pi }{6}$

$A=\frac{\pi }{6}$

$\text{Hence,}\text{this}\text{is the answer}\text{.}$