If in the ∆ABC, ∠B=45°, then a4+b4+c4 is equal to:
2a2b2+c2
2c2a2+b2
2b2a2+c2
2a2b2+b2c2+c2a2
Explanation for the correct option:
Find the value of a4+b4+c4:
It is given that ∠B=45°, by applying the cosine formula, we get
b2=a2+c2-2accosB⇒2accos45°=a2+c2-b2⇒2ac12=a2+c2-b2
By squaring both sides, we get
4a2c22=a2+c2-b22⇒2a2c2=a4+b4+c4+2a2c2-2b2c2-2a2b2⇒a4+b4+c4=2b2c2+2a2b2⇒a4+b4+c4=2b2c2+a2
Hence, option C is correct.
If in the triangle ABC, B=450,then a4+b4+c4 is equal to