Question

Answer 1

\(BM=2CM=2\left(BC-BM\right)\Rightarrow3BM=2BC\Rightarrow BM=\dfrac{2}{3}BC=4\)

\(cosA=\dfrac{AB^2+AC^2-BC^2}{2AB.AC}=\dfrac{43}{48}\)

\(\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cosA=86\)

\(cosB=\dfrac{AB^2+BC^2-AC^2}{2AB.BC}=\dfrac{29}{36}\)

\(\overrightarrow{BM}.\overrightarrow{BA}=BM.AB.cosB=\dfrac{116}{3}\)

\(AM=\sqrt{AB^2+BM^2-2AB.BM.cosB}=\dfrac{2\sqrt{186}}{3}\)