Đáp án:
$\begin{array}{l}
\Delta ABC \bot A;AH \bot BC\\
\Rightarrow \widehat C = \widehat {BAH}\left( { + \widehat B = {{90}^0}} \right)\\
Trong\Delta ABH \bot H\\
\Rightarrow B{H^2} = A{B^2} - A{H^2} = {30^2} - {24^2}\\
\Rightarrow BH = 18\left( {cm} \right)\\
\Rightarrow \left\{ \begin{array}{l}
\sin \widehat C = \sin \widehat {BAH} = \dfrac{{BH}}{{AB}} = \dfrac{{18}}{{30}} = \dfrac{3}{5}\\
cos\widehat C = cos\widehat {BAH} = \dfrac{{AH}}{{AB}} = \dfrac{{24}}{{30}} = \dfrac{4}{5}\\
\tan \widehat C = \dfrac{3}{4}\\
\cot \widehat C = \dfrac{4}{3}
\end{array} \right.
\end{array}$
