Check the picture below.
[tex]\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \sqrt{6^2 - 4^2}=a\implies \sqrt{36-16}=a\implies \sqrt{20}=a\implies 2\sqrt{5}=a \\\\[-0.35em] ~\dotfill\\\\ sin(A )=\cfrac{\stackrel{opposite}{2\sqrt{5}}}{\underset{hypotenuse}{6}}\implies sin(A)=\cfrac{\sqrt{5}}{3} \\\\\\ cos(B )=\cfrac{\stackrel{adjacent}{2\sqrt{5}}}{\underset{hypotenuse}{6}}\implies cos(B)=\cfrac{\sqrt{5}}{3}[/tex]
