In\(\Delta\) ABC, right-angled at C, sin A = $ \frac{5}{13}$, then COS B?

In triangle ABC, the triangle is right angles at C. Therefore, the two acute angles A and B are complementary.
A + B = 90 Degree

$\sin A=\frac{5}{13},\ A+B=90^\circ \cos B=\sin A=\frac{5}{13}$