In a triangle ABC,if 2a^2b^2 + 2b^2c^2 = a^4 | vedclass

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(A) Given the equation: $2a^2b^2 + 2b^2c^2 = a^4 + b^4 + c^4$ Rearranging the terms: $a^4 + b^4 + c^4 - 2a^2b^2 - 2b^2c^2 = 0$ We know the identity: $

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$45^o$ or $135^o$

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(A) Given the equation: $2a^2b^2 + 2b^2c^2 = a^4 + b^4 + c^4$ Rearranging the terms: $a^4 + b^4 + c^4 - 2a^2b^2 - 2b^2c^2 = 0$ We know the identity: $(a^2 - b^2 + c^2)^2 = a^4 + b^4 + c^4 - 2a^2b^2 + 2a^2c^2 - 2b^2c^2$ Substituting the given condition into the identity: $(a^2 - b^2 + c^2)^2 = 2a^2c^2$ Taking the square root on both sides: $a^2 - b^2 + c^2 = \pm \sqrt{2}ac$ Dividing by $2ac$: $\frac{a^2 + c^2 - b^2}{2ac} = \pm \frac{\sqrt{2}ac}{2ac} = \pm \frac{1}{\sqrt{2}}$ By the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,so $\cos B = \pm \frac{1}{\sqrt{2}}$ Thus,$B = 45^o$ or $B = 135^o$.

In a triangle $ABC$,if $2a^2b^2 + 2b^2c^2 = a^4 + b^4 + c^4$,then angle $B$ is equal to:

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