In Delta ABC, if (a + b + c)(a - b + c) = 3ac | vedclass

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(A) Given: $(a + c + b)(a + c - b) = 3ac$Using the identity $(x+y)(x-y) = x^2 - y^2,$ we get:$(a + c)^2 - b^2 = 3ac$$a^2 + c^2 + 2ac - b^2 = 3ac$$a^2

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$\angle B = 60^\circ$

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(A) Given: $(a + c + b)(a + c - b) = 3ac$Using the identity $(x+y)(x-y) = x^2 - y^2,$ we get:$(a + c)^2 - b^2 = 3ac$$a^2 + c^2 + 2ac - b^2 = 3ac$$a^2 + c^2 - b^2 = ac$Dividing both sides by $2ac,$ we get:$\frac{a^2 + c^2 - b^2}{2ac} = \frac{ac}{2ac} = \frac{1}{2}$By the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}.$Therefore,$\cos B = \frac{1}{2}.$Since $\cos 60^\circ = \frac{1}{2},$ we have $\angle B = 60^\circ.$

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

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