In $Δ A B C, i f (a + b + c) (a - b + c) = 3 a c$ , then

∠ B = 60^{\circ}

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∠ B = 30^{\circ}

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∠ C = 60^{\circ}

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∠ A + ∠ B = 120^{\circ}

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Solution

The correct option is A $∠ B = 60^{\circ}$
$(a + c)^{2} - b^{2} = 3 a c$
$\Rightarrow a^{2} + c^{2} - b^{2} = a c$
Applying cosine rule we get,
$\Rightarrow c o s B = \frac{1}{2}$
$\Rightarrow B = 60^{\circ}$

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