Đáp án:
$\begin{array}{l}
\widehat {BAH} + \widehat B = {90^0}\\
\widehat C + \widehat B = {90^0}\\
\Rightarrow \widehat {BAH} = \widehat C\\
\Rightarrow \sin \widehat C = \sin \widehat {BAH} = \dfrac{2}{5}\\
\Rightarrow \dfrac{2}{5} = \dfrac{{AB}}{{BC}}\\
\Rightarrow BC = \dfrac{5}{2}AB = 15\left( {cm} \right)\\
\Rightarrow AC = \sqrt {{{15}^2} - {6^2}} = 3\sqrt {21} \left( {cm} \right)\\
Do:sin\widehat C = \dfrac{2}{5} \Rightarrow \widehat C = {23^0}\\
\Rightarrow \widehat B = {90^0} - {23^0} = {67^0}\\
Vậy\,AC = 3\sqrt {21} cm;BC = 15cm;\\
\widehat C = {23^0};\widehat B = {67^0}
\end{array}$
