In a right-angled triangle, one of the acute angles measures 53°.

Let ABC be a triangle right-angled at B.

Then, $\angle B=90^{\circ}$ and let $\angle A=53^{\circ}$.

$\therefore \angle A+\angle B+\angle C=180^{\circ} \quad$ [Sum of the angles of a triangle]

$\Rightarrow 53^{\circ}+90^{\circ}+\angle C=180^{\circ}$

$\Rightarrow \angle C=37^{\circ}$

Hence, $\angle A=53^{\circ}, \angle B=90^{\circ}$ and $\angle C=37^{\circ}$.