Đáp án:
$\begin{array}{l}
+ \widehat B + \widehat C = {90^0}\\
\Rightarrow \widehat C = {30^0}\\
Do:\cos \widehat B = \dfrac{{AB}}{{BC}}\\
\Rightarrow \cos {60^0} = \dfrac{1}{2} = \dfrac{{6\sqrt 3 }}{{BC}}\\
\Rightarrow BC = 12\sqrt 3 \left( {cm} \right)\\
\sin \widehat B = \dfrac{{AC}}{{BC}}\\
\Rightarrow AC = \sin {30^0}.12\sqrt 3 = \dfrac{{\sqrt 3 }}{2}.12\sqrt 3 = 18\left( {cm} \right)\\
Vay\,AC = 18\left( {cm} \right);\\
BC = 12\sqrt 3 \left( {cm} \right);\widehat C = {30^0}
\end{array}$
