Đáp án:
Trong tam giác ABH vuông tại H có:
$\begin{array}{l}
\sin \widehat B = \dfrac{{AH}}{{AB}}\\
\Rightarrow AH = \sin {57^0}.9 = 7,55\\
\Rightarrow BH = \sqrt {A{B^2} - A{H^2}} = 4,9\\
Trong\,\Delta AHC \bot tai\,H\\
\Rightarrow HC = \sqrt {A{C^2} - A{H^2}} = 9,33\\
\Rightarrow BC = HB + HC = 14,23\\
\Rightarrow \sin \widehat C = \dfrac{{AH}}{{AC}} = 0,63 \Rightarrow \widehat C = {39^0}\\
\Rightarrow \widehat A = {180^0} - \widehat C - \widehat B = {84^0}\\
+ )Do:BM = \dfrac{{BC}}{2} = 7,115\\
\Rightarrow HM = BM - BH = 2,215\\
\Rightarrow AM = \sqrt {A{H^2} + H{M^2}} = 7,868
\end{array}$