\(\begin{array}{l}
1B\\
2B\\
3B.\\
AB = BC.\sin \widehat C = 12.\sin {60^0} = 12.\frac{{\sqrt 3 }}{2} = 6\sqrt 3 \\
AC = BC.\cos \widehat C = 12.\cos {60^0} = 12.\frac{1}{2} = 6\\
4.A\\
AH.BC = AB.AC \Leftrightarrow AH = \frac{{AB.AC}}{{BC}} = \frac{{6.6\sqrt 3 }}{{12}} = 3\sqrt 3 \\
5.C\\
\cot B = \frac{{AB}}{{AC}} = \frac{{6\sqrt 3 }}{6} = \sqrt 3 \\
6.D\\
Ke\,\,AH \bot BC \Rightarrow HB = HC = \frac{{BC}}{2}\\
\widehat {BAH} = \frac{{\widehat {BAC}}}{2} = \frac{{{{120}^0}}}{2} = {60^o}\\
Xet\,\Delta AHB\,co:\,HB = AB.\sin {60^0} = 6.\frac{{\sqrt 3 }}{2} = 3\sqrt 3 \\
\Rightarrow BC = 2HB = 6\sqrt 3
\end{array}\)
