Question

In a ∆ABC, if a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c², then B = __________.

Answer 1

In a ∆ABC,
Given a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c²
i.e a⁴ + b⁴ − 2a²b² + c² − 2b²c² = 0
i.e (a² + c²)² + b⁴ − 2a²b² − 2b²c² − 2a²c² = 0
i.e (a² + c²)² + b⁴ − 2b² (a² + c²) − 2a²c² = 0
i.e (a² + c² − b²)² = 2a²c²
i.e a² + c² − b² = $\sqrt{2}ac$
Using cosine formula,
i.e cosB = $\frac{a^2+c^2-b^2}{2ac}=\frac{\sqrt{2}ac}{2ac}$
cosB = $\frac{1}{\sqrt{2}}$
i.e $B=\frac{\mathrm{\pi }}{4}\mathrm{or}\mathrm{\pi }-\frac{\mathrm{\pi }}{4}=\frac{3\mathrm{\pi }}{4}$