Question

If a, b and c are the sides of a triangle such that a⁴ + b⁴ + c⁴ = 2c² (a² + b²) then the angles opposite
to the side C is:
(A) 45° or 135° (B) 30° or 100° (C) 50° or 100° (D) 60° or 120°

Answer 1

$$\begin{gathered} a^4+b^4+c^4=2c^2\left(a^2+b^2\right) \\ {a}^4+b^4+c^4=2c^2{a}^2+2c^2{b}^2 \\ {a}^4+b^4+c^4-2c^2{a}^2-2c^2{b}^2=0 \\ {a}^4+b^4+c^4-2c^2{a}^2-2c^2{b}^2+2a^2{b}^2=2a^2{b}^2 \\ {\left(a^2\right)}^2+{\left(b^2\right)}^2+{\left(-c^2\right)}^2+2\left(-c^2\right)a^2+2\left(-c^2\right)b^2+2a^2{b}^2=2a^2{b}^2 \\ {\left(a^2+b^2-c^2\right)}^2=2a^2{b}^2 \\ \left(a^2+b^2-c^2\right)=\pm \sqrt{2a^2{b}^2} \\ {\left(a^2+b^2-c^2\right)}^2=\pm \sqrt{2}ab \\ By\cos inerule \\ CosC=\frac{a^2+b^2-c^2}{2ab}=\frac{\pm \sqrt{2}ab}{2ab}=\pm \frac{1}{\sqrt{2}} \\ CosC=\frac{1}{\sqrt{2}}or\cos C=-\frac{1}{\sqrt{2}} \\ CosC=Cos45°or\cos C=-\cos 45° \\ CosC=Cos45°or\cos C=\cos \left(180°-45°\right) \\ C=45°orC=180°-45°=135° \end{gathered}$$