MCQ
In $\triangle A B C$, if $(a+b+c)(a-b+c)=3 a c$, then
  • $\angle B =60^{\circ}$
  • B
    $\angle B =30^{\circ}$
  • C
    $\angle C =60^{\circ}$
  • D
    $\angle A +\angle C =90^{\circ}$

Answer

Correct option: A.
$\angle B =60^{\circ}$
(A) $(a+b+c)(a-b+c)=3 a c$
$\begin{array}{l}\Rightarrow a^2+2 a c+c^2-b^2=3 a c \\ \Rightarrow a^2+c^2-b^2=a c\end{array}$
But $\cos B =\frac{ a ^2+ c ^2- b ^2}{2 ac }=\frac{1}{2} \Rightarrow B=60^{\circ}$

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