Question

In a triangle ABC,
${c}^{4} - 2( {a}^{2} + {b}^{2} ) {c}^{2} + {a}^{4} + {a}^{2} {b}^{2} + {b}^{4} = 0$
Then the acute angle c = ?​

Answer 1

Step-by-step explanation:

I think your question is not proper

...

Answer 2

Answer:

It is given that 2a2+4b2+c2=4ab+2ac

Simplifying, we get 

(a2−2ac+c2)+(a2−4ab+4b2)=0

(a−c)2+(a−2b)2=0

a=c and a=2b

Hence

cosB=2aca2+c2−b2

=2a22a2−b2

=8b28b2−b2

=87.