Question

In $\Delta ABC$
$(a+b+c)(a-b+c)=3ac,$ then find $\angle B$ in degree

Answer 1

$(a+b+c)(a-b+c)=3ac,$

simplifying

or $(a+c{)}^2-b^2=3ac$

or $a^2+c^2-b^2=ac$

$\therefore \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{1}{2}$
or

$\angle B=\frac{\pi }{3}={60}^0$