\(S=\dfrac{1}{2}\cdot AC\cdot AB\cdot\sin60^0=10\sqrt{3}\)
Xét ΔBAC có \(\cos A=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
\(\Leftrightarrow\dfrac{89-BC^2}{80}=\dfrac{1}{2}\)
\(\Leftrightarrow BC=7\)
\(\Leftrightarrow AH\cdot BC=20\sqrt{3}\)
hay \(AH=\dfrac{20\sqrt{3}}{7}\)