In triangle $A B C$ if $2 a^{2} b^{2} + 2 b^{2} c^{2} = a^{4} + b^{4} + c^{4},$ then angle $B$ is equal to

45^{\circ}

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135^{\circ}

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120^{\circ}

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60^{\circ}

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Solution

The correct options are
C $45^{\circ}$
D $135^{\circ}$
$2 a^{2} b^{2} + 2 b^{2} c^{2} = a^{4} + b^{4} + c^{4}$
$a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 b^{2} c^{2} + 2 c^{2} a^{2} = 2 c^{2} a^{2}$
$(c^{2} + a^{2} - b^{2})^{2} = 2 c^{2} a^{2} \Rightarrow c^{2} + a^{2} - b^{2} = \pm \sqrt{2} c a$
Thus using cosine rule $cos B = \frac{c^{2} + a^{2} - b^{2}}{2 a c} = \frac{\pm \sqrt{2} c a}{2 c a} = \pm \frac{1}{\sqrt{2}} \Rightarrow ∠ B = 45^{o} or 135^{o}$

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Similar questions

If in the triangle ABC, $B = 45^{0}, t h e n a^{4} + b^{4} + c^{4}$ is equal to

a^{4} - 2 (b^{2} + c^{2}) a^{2} + b^{4} + b^{2} c^{2} + c^{4} = 0

∠ A

60^{\circ}

120^{\circ}

If $a + b + c = 0$ then prove

$a^{4} + b^{4} + c^{4} = 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})$

△ A B C, a^{4} + b^{4} + c^{4} = 2 a^{2} b^{2} + b^{2} c^{2} + 2 c^{2} a^{2},

sin A

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