Write the measures of the sides of $\Delta PQR$ in ascending order,given that $\angle P = 40^{\circ}$ and $\angle R = 80^{\circ}$.

(QR, PR, PQ) In $\Delta PQR$,the sum of all interior angles is $180^{\circ}$.
Therefore,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the given values: $40^{\circ} + \angle Q + 80^{\circ} = 180^{\circ}$.
$120^{\circ} + \angle Q = 180^{\circ}$,which gives $\angle Q = 60^{\circ}$.
We know that the side opposite to the smallest angle is the shortest,and the side opposite to the largest angle is the longest.
The angles in ascending order are $\angle P (40^{\circ}) < \angle Q (60^{\circ}) < \angle R (80^{\circ})$.
Therefore,the sides opposite to these angles in ascending order are $QR < PR < PQ$.