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 the angles opposite to the side $c$ is

• A. $45°$or $90°$
• B. $30°$or $135°$
• C. $45°$or $135°$
• D. $60°$or $120°$

Answer 1

The correct option is C

$45°$or $135°$

Explanation for the correct option :

Step1. Draw the diagram:

Let $C$ be the angle opposite to the side $c$.

Given $a,b$ and $c$ are the sides of a triangle such that $a^4+b^4+c^4=2c^2(a^2+b^2)$

$\Rightarrow a^4+b^4+c^4-2c^2{a}^2-2c^2{b}^2=0$

Add $2a^2{b}^2$ on both sides

$$\begin{gathered} \Rightarrow a^4+b^4+c^4–2c^2{a}^2–2c^2{b}^2+2a^2{b}^2=2a^2{b}^2 \\ \Rightarrow {(a^2+b^2–c^2)}^2=2(ab{)}^2 \end{gathered}$$

Step2. Finding the angles opposite to the side $c$:

Take square root on both sides

$$\begin{gathered} \Rightarrow (a^2+b^2–c^2)=\pm \sqrt{2}\left(ab\right) \\ \Rightarrow \frac{(a^2+b^2–c^2)}{2ab}=\frac{\pm \sqrt{2}}{2} \\ \Rightarrow \cos C=\pm \frac{1}{\sqrt{2}}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{C}\mathbf{=}\frac{\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}\mathbf{}{\mathbf{b}}^{\mathbf{2}}\mathbf{–}\mathbf{}{\mathbf{c}}^{\mathbf{2}}\mathbf{)}\mathbf{}}{\mathbf{2}\mathbf{a}\mathbf{b}}\mathbf{]} \end{gathered}$$

$\mathbf{\therefore }\mathbf{\angle }\mathit{C}\mathbf{}\mathbf{=}\mathbf{}\mathbf{45}\mathbf{°}$ or $\mathbf{135}\mathbf{°}$

Hence, the correct option is C.