Question

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

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

Answer 1

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

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

$\frac{b(a+b)+c(a+c)}{(a+b)(a+c)}=1$

$ab+b^2+ac+c^2=a^2+ac+ab+1$

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

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

By Cosine Rule,

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

$\therefore \cos A=1/2A={60}^o$