trig problem

Here's my attempt (I'm not good at trig): 

C

5             7

A                           B

Let $AC = 5$ and $BC = 7$. 

By the Pythagorean Theorem, $AB = \sqrt{24} = 2\sqrt6$

So, $\cos B = \color{brown}\boxed{2 \sqrt 6 \over 7}$

Here's another way

cos B  =  sqrt (1 - sin^2B)  =  sqrt ( 1 - 25/49 )  =  sqrt ( 49 -25) / 7  =  sqrt ( 24)  / 7  =   

2sqrt (6)  /  7        (as BuilderBoi found !!!)

$In\;the\; \Delta ABC, we\; have\; the \;\angle A\;=90^{\circ}.\;sin(B)\;=\; \frac{5}{7}.\;Find\;cos(B).$

Sorry guys, I kinda f'ed up cos and sin in the drawing, but the calculation is correct.