Question

In $△ABC$ if $2a^2{b}^2+2b^2{c}^2=a^4+b^4+c^4,\angle B=$

• A. ${45}^0$ or ${135}^0$
• B. ${135}^0$ or ${120}^0$
• C. ${30}^0$ or ${60}^0$
• D. ${40}^0$ or ${60}^0$

Answer 1

The correct option is A ${45}^0$ or ${135}^0$

We know that,

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

$\Rightarrow {(a^2-b^2+c^2)}^2=2a^2{b}^2+2b^2{c}^2-2a^2{b}^2-2b^2{c}^2+2a^2{c}^2$

$\therefore {(a^2-b^2+c^2)}^2=2a^2{c}^2$

$\Rightarrow \left(\frac{a^2+b^2-c^2}{2ac}\right)=\frac{1}{2}$

$\Rightarrow \frac{a^2+c^2-b^2}{2ac}=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \cos B=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow B={45}^0$ or ${135}^0$