Respuesta :
In [tex] \Delta ABC [/tex], [tex] C=90^o [/tex]. Therefore,
[tex] A+B=90^o\\
B=90^o-A [/tex].
Also [tex] \cos B=\frac{5}{13} \\
\cos (90^o-A)=\frac{5}{13} \\
\sin A=\frac{5}{13} \\ [/tex].
Thus, [tex] \sin A [/tex] equals the given fraction.