[tex]A+B=90^o\to B=90^o-A\\\\\cos B=\cos(90^o-A)=\cos90^o\cos A+\sin90^o\sin A\\\\=\cos A+1\cdot\dfrac{3}{5}=0+\dfrac{3}{5}=\dfrac{3}{5}\\\\\boxed{\cos B=\dfrac{3}{5}}[/tex]
[tex]Used:\\\\\cos(x+y)=\cos x\cos y+\sin x\sin y\\\\\cos90^o=0,\ sin90^o=1[/tex]
Other method:
Look at the picture.
[tex]\sin\theta=\dfrac{opposite}{hypotenuse}\\\\\sin A=\dfrac{3}{5}\to opposite=3\ and\ hypotenuse=5[/tex]
[tex]\cos\theta=\dfrac{adjacent}{hypotenuse}\\\\adjacent=3\ and\ hypotenuse=5\to\cos B=\dfrac{3}{5}[/tex]
