Question

In ΔABC, what is the value of cos B?

Answer 1

`b^(2) = a^(2) + c^(2) -2acCosB`
Making cos B the subject;
2acCosB=
`a^(2) + c^(2) - b^(2)`
cosB=
`( a^(2) + c^(2) - b^(2) )/(2ac)`