Question

If $a,b$ and $c$ are the sides of a triangle such that $a^4+b^4+c^4=2c^2(a^2+b^2),$ then find the possible angles opposite to the side $c.$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{4}$
• D. $\frac{3\pi }{2}$

Answer 1

Answer: (A), (C)

$\frac{\pi }{4}$
$\frac{3\pi }{4}$

Given, $a^4+b^4+c^4=2c^2(a^2+b^2)$ ...(1)

Now, as $\cos C=\frac{a^2+b^2-c^2}{2ab}$

$\Rightarrow {\cos }^{2}C=\frac{a^4+b^4+c^4+2a^2{b}^2-2b^2{c}^2-2a^2{c}^2}{4a^2{b}^2}$ (using (1))

$\Rightarrow \cos C=\pm \frac{1}{\sqrt{2}}\Rightarrow C=\frac{\pi }{4}$ or $\frac{3\pi }{4}$