Given PQ || RT, QR || SU, angle PQR = 48 degrees, and angle RTS = x, find the value of x.

We are given that $PQ \parallel RT$ and $QR \parallel SU$.
We are given $\angle PQR = 48^\circ$ and $\angle RTS = x$.

Consider the parallel lines $PQ$ and $RT$ intersected by the transversal $QR$.
The angles $\angle PQR$ and $\angle QRT$ are alternate interior angles.
Therefore, they are equal:

Consider the parallel lines $QR$ and $SU$ intersected by the transversal $RT$.
The angles $\angle QRT$ and $\angle RTS$ are consecutive interior angles (also known as co-interior angles).
The sum of consecutive interior angles is $180^\circ$.
So, we have:

Substitute $\angle QRT = 48^\circ$ and $\angle RTS = x$ into the equation:

Subtract $48^\circ$ from both sides:

The value of $x$ is $\boxed{\text{132}^\circ}$.