Question

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

• A. $A={60}^0$
• B. $B={60}^0$
• C. $C={60}^0$
• D. $A={90}^o$

Answer 1

The correct option is A $A={60}^0$

$\frac{a+c+a+b}{(a+b)(a+c)}=\frac{3}{(a+b+c)}$
$(2a+b+c)(a+b+c)=3(a^2+ab+ac+bc)$
$2ab+2ac+2a^2+ab+ac+b^2+c^2+2bc=3a^2+3ab+3ac+3bc$ ...(on solving this equation we get)
$\frac{b^2+c^2-a^2}{bc}=1$
$\cos A=\frac{b^2+c^2-a^2}{2bc}$... (cos formula)
$\cos A=\frac{1}{2}$
$A=\pi /3$
$={60}^0$.