Question

In $△$ ABC, if (a+b+c)(a-b+c)=3ac, then
[AMU 1996]

• A. $\angle B={60}^o$
• B. $\angle B={30}^o$
• C. $\angle C={60}^o$
• D. $\angle A+\angle C={90}^o$

Answer 1

The correct option is A

$\angle B={60}^o$

$(a+c{)}^2-b^2=3ac\Rightarrow a^2+c^2-b^2=ac$
But cos $B=\frac{a^2+c^2-b^2}{2ac}=\frac{1}{2}\Rightarrow B=\frac{\pi }{3}.$