Question

In $\Delta ABC$, if $c^4-2(a^2+b^2)c^2+a^4+a^2{b}^2+b^4=0$ then $\angle C=$

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${75}^0$

Answer 1

The correct option is C ${60}^0$

We know that In $\Delta ABC$
$\cos C=\frac{a^2+b^2-c^2}{2ab}$
Now,
$c^4-2(a^2+b^2)c^2+a^4+2a^2{b}^2+b^4=a^2{b}^2$
$\Rightarrow {(a^2+b^2-c^2)}^2=a^2{b}^2$
$\Rightarrow {\left(\frac{a^2+b^2-c^2}{ab}\right)}^2=1$
$\Rightarrow {\left(\frac{a^2+b^2-c^2}{ab}\right)}^{}=1or{\left(\frac{a^2+b^2-c^2}{ab}\right)}^{}=-1$
$\Rightarrow \cos C=\frac{1}{2}$ or $\cos C=\frac{-1}{2}$
$\therefore C={60}^0$ or $C={120}^0$