notice those two pictures below, those are two examples with a right-triangle correspondences.
therefore, in this case, we can simply nevermind the 90° angle and instead use the other two, A and C.
[tex]\bf cos(C)=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}} \qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{3}}{\stackrel{hypotenuse}{5}} \qquad cos(C)=sin(A)
\\\\\\
sin(C)=\cfrac{\stackrel{opposite}{4}}{\stackrel{hypotenuse}{5}} \qquad \qquad cos(A)=\cfrac{\stackrel{adjacent}{4}}{\stackrel{hypotenuse}{5}} \qquad sin(C)=cos(A)[/tex]
