Since the magnitudes satisfy the identity for right triangles
[tex]A^2=B^2+C^2[/tex]
we deduce that B and C are perpendicular, and that ABC is a right triangle (see picture).
So, we can get the angle between A and C using the sine theorem:
[tex]\dfrac{A}{\sin(\hat{A})}=\dfrac{B}{\sin(\hat{B})}[/tex]
And we have
[tex]A=5,\quad B=4,\quad \hat{A}=90[/tex]
So, we have
[tex]\sin(\hat{B})=\dfrac{B\sin(\hat{A})}{A}=\dfrac{4\sin(90)}{5}=\dfrac{4}{5}[/tex]
Which implies
[tex]\hat{B}=\arcsin\left(\dfrac{4}{5}\right)\approx 53.13[/tex]