Given a right angle triangle ABC:
[tex]\begin{gathered} m\angle C=90 \\ \sin A=\frac{3}{5} \end{gathered}[/tex]As the measure of angle C = 90
so, the sum of the angles A and B = 90
So, angles A and B are complementary angles
so,
[tex]\begin{gathered} \sin A=\cos B \\ \\ \cos B=\sin A=\frac{3}{5} \end{gathered}[/tex]so, the answer will be cos B = 3/5
