In ΔABC, what is the value of cos B?
\(b^{2} = a^{2} + c^{2} -2acCosB\)Making cos B the subject;2acCosB=\(a^{2} + c^{2} - b^{2}\)cosB=\(\frac{ a^{2} + c^{2} - b^{2} }{2ac}\)
