$\because$将$\triangle ABC$绕点$B$逆时针旋转$60^{\circ}$得到$\triangle {A\’}BC\’$，
$\therefore AB={A\’}B$，$\angle ABA\’=60^{\circ}$，$BC=BC\’=3$，
$\therefore \triangle ABA\’$是等边三角形，
$\therefore AB=AA\’$，
$\because \angle C=90^{\circ}$，
$\therefore AB=\sqrt{A{C}^{2}+B{C}^{2}}=\sqrt{9+16}=5$，
$\therefore AA\’=AB=5$.

$\because$将$\triangle ABC$绕点$B$逆时针旋转$60^{\circ}$得到$\triangle {A\’}BC\’$，
$\therefore AB={A\’}B$，$\angle ABA\’=60^{\circ}$，$BC=BC\’=3$，
$\therefore \triangle ABA\’$是等边三角形，
$\therefore AB=AA\’$，
$\because \angle C=90^{\circ}$，
$\therefore AB=\sqrt{A{C}^{2}+B{C}^{2}}=\sqrt{9+16}=5$，
$\therefore AA\’=AB=5$.