Lời giải
Ta có \(B{C^2} = A{B^2} + A{C^2} - 2.AB.AC.\cos \widehat {BAC}\)
= 32 + 52 – 2.3.5.cos60°
= 19.
Suy ra \(BC = \sqrt {19} \).
Khi đó \[\cos B = \frac{{A{B^2} + B{C^2} - A{C^2}}}{{2.AB.BC}} = \frac{{{3^2} + {{\left( {\sqrt {19} } \right)}^2} - {5^2}}}{{2.3.\sqrt {19} }} = \frac{{\sqrt {19} }}{{38}}\].
Ta có BM = 2MC.
\( \Rightarrow BM = \frac{2}{3}BC = \frac{{2\sqrt {19} }}{3}\).
Do đó AM2 = AB2 + BM2 – 2.AB.BM.cosB
\( = {3^2} + {\left( {\frac{{2\sqrt {19} }}{3}} \right)^2} - 2.3.\frac{{2\sqrt {19} }}{3}.\frac{{\sqrt {19} }}{{38}} = \frac{{139}}{9}\).
Vậy \(AM = \frac{{\sqrt {139} }}{3}\).