Đáp án:
Trong tam giác ABC vuông tại A, theo Pytago ta có:
$\begin{array}{l}
BC = \sqrt {A{B^2} + A{C^2}} = \sqrt {{6^2} + {8^2}} = 10\\
\Rightarrow \left\{ \begin{array}{l}
\sin \widehat B = \cos \widehat C = \dfrac{{AC}}{{BC}} = \dfrac{8}{{10}} = \dfrac{4}{5}\\
\cos \widehat B = \sin \widehat C = \dfrac{{AB}}{{BC}} = \dfrac{6}{{10}} = \dfrac{3}{5}\\
\tan \widehat B = \cot \widehat C = \dfrac{{AC}}{{AB}} = \dfrac{4}{3}\\
\cot \widehat B = \tan \widehat C = \dfrac{3}{4}
\end{array} \right.
\end{array}$
