Question

In $\Delta ABC,if(a+b+c)(a-b+c)=3ac$, then

• A. $\angle B={60}^{\circ }$
• B. $\angle B={30}^{\circ }$
• C. $\angle C={60}^{\circ }$
• D. $\angle A+\angle B={120}^{\circ }$

Answer 1

The correct option is A $\angle B={60}^{\circ }$

$(a+c{)}^2-b^2=3ac$
$\Rightarrow a^2+c^2-b^2=ac$
Applying cosine rule we get,
$\Rightarrow cosB=\frac{1}{2}$
$\Rightarrow B={60}^{\circ }$