Q68 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics

In a $\triangle A B C, a^4+b^4+c^4=2 b^2 c^2+2 a^2 b^2$, then $B=$

  • Correct answer
    A
    $\frac{\pi}{4}$ or $\frac{3 \pi}{4}$
  • B
    $\frac{\pi}{2}$
  • C
    $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$
  • D
    $\frac{\pi}{6}$ or $\frac{5 \pi}{6}$

Solution

We have, $\quad a^4+b^4+c^4=2 b^2\left(c^2+a^2\right)$ $$ \Rightarrow a^4+b^4+c^4-2 b^2 c^2-2 a^2 b^2=0 $$ Add $2 a^2 c^2$ on both sides, we get $$ \begin{array}{ll} \Rightarrow a^4+b^4+c^4-2 b^2 c^2-2 a^2 b^2+2 c^2 a^2=2 c^2 a^2 \\ \Rightarrow \quad\left(a^2+c^2-b^2\right)^2=2 c^2 a^2 \\ \Rightarrow & \frac{4\left(a^2+c^2-b^2\right)^2}{(2 c a)^2}=2 \\ \Rightarrow & \frac{\left(a^2+c^2-b^2\right)^2}{(2 c a)^2}=\frac{1}{2} \quad\left[\because \cos B=\frac{c^2+a^2-b^2}{2 c a}\right] \\ \therefore & \cos ^2 B=\frac{1}{2} \\ \Rightarrow & \cos B=\frac{1}{\sqrt{2}} \end{array} $$ So, we get $\quad \angle C=\frac{\pi}{4}$ or $\frac{3 \pi}{4}$

Hence, the correct answer is option: 'A'.

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