Đáp án:
$\begin{array}{l}
1)Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2} = {8^2} + {15^2} = 289\\
\Rightarrow BC = 17\\
\tan \widehat B = \dfrac{{AC}}{{AB}} = \dfrac{{15}}{8}\\
\cot \widehat B = \dfrac{8}{{15}}\\
\sin \widehat B = \dfrac{{AC}}{{BC}} = \dfrac{{15}}{{17}}\\
\cos \widehat B = \dfrac{8}{{17}}\\
B2)\\
\cot \widehat B = 2,4 = \dfrac{{12}}{5} = \dfrac{{AB}}{{AC}}\\
\Rightarrow \tan \widehat B = \dfrac{5}{{12}}\\
\Rightarrow AC = 2,5\\
\Rightarrow BC = \sqrt {{6^2} + 2,{5^2}} = 6,5\\
\Rightarrow \sin \widehat B = \dfrac{{AC}}{{BC}} = \dfrac{{2,5}}{{6,5}} = \dfrac{5}{{13}}\\
\cos \widehat B = \dfrac{{AB}}{{BC}} = \dfrac{6}{{6,5}} = \dfrac{{12}}{{13}}\\
B3)BC = \sqrt {{6^2} + {8^2}} = 10\\
+ \sin \widehat B = \cos \widehat C = \dfrac{{AC}}{{BC}} = \dfrac{8}{{10}} = \dfrac{4}{5}\\
+ \cos \widehat B = \sin \widehat C = \dfrac{3}{5}\\
+ \tan \widehat B = \cot \widehat C = \dfrac{4}{3}\\
+ \cot \widehat B = \tan \widehat C = \dfrac{3}{4}
\end{array}$
