In a triangle ABC if $a^{4} - 2 (b^{2} + c^{2}) a^{2} + b^{4} + b^{2} c^{2} + c^{4} = 0$ then show $∠ A$ is $60^{\circ}$ or $120^{\circ}$

Open in App

Solution

$a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 a^{2} c^{2} + b^{2} c^{2} = 0$
$a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 a^{2} c^{2} + b^{2} c^{2} + b^{2} c^{2} = b^{2} c^{2}$
${(a^{2} - b^{2} - c^{2})}^{2} = b^{2} c^{2} \Rightarrow a^{2} - b^{2} - c^{2} = \pm b c$
weknow $c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \Rightarrow cos A = \pm \frac{1}{2} \Rightarrow A = 60 o r 120$

Suggest Corrections

Similar questions

If $a + b + c = 0$ then prove

$a^{4} + b^{4} + c^{4} = 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})$

a + b + c = 0

\frac{a^{4} + b^{4} + c^{4}}{a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}}

a^{4} + b^{4} + c^{4} \geq a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}

Δ A B C

c^{4} - 2 (a^{2} + b^{2}) c^{2} + a^{4} + a^{2} b^{2} + b^{4} = 0

∠ C =

△ A B C

2 a^{2} b^{2} + 2 b^{2} c^{2} = a^{4} + b^{4} + c^{4}, ∠ B =

Join BYJU'S Learning Program