From right [tex]\triangle ABC,[/tex] we have [tex]\sin B=\frac{AC}{AB}=\frac{2}{3}.[/tex] Using a calculator, we find [tex]\angle B=\arcsin\frac{2}{3}\approx\boxed{41.81^\circ}.[/tex] Since [tex]\angle B[/tex] is acute, this answer is correct, as this lies within the range of the arcsine function.
