Find the angle between the sides 4 and 7 in a right triangle

Question

I need to solve the $B$ corner

What I've tried:

$$\operatorname{sin} B=\frac47$$ $$B=\operatorname{arcsin}\frac47$$ $$B=34.85$$

But that's not the right answer, can anyone help me find what I did wrong?

Answer 1

It should be $\cos B=\frac 47$

Answer 2

$\sin B\neq \frac{4}{7}$

$\sin B=\frac{AC}{7}$

$AC=\sqrt{33}$ by pythagorean theorem.