In the figure, PQRS is a rhombus and STUV is a parallelogram. PSV is a straight line. $\angle PSQ = 48^\circ$, $\angle RTS = 21^\circ$ and $\angle STU = 116^\circ$.

1. Identify the isosceles triangle formed by the sides of the rhombus to find the base angles

Since $PQRS$ is a rhombus, all its sides are equal, so $PQ = PS$.
This makes $\triangle PQS$ an isosceles triangle.
Therefore, $\angle PQS = \angle PSQ = 48^\circ$.

2. Calculate $\angle QPS$ using the angle sum of a triangle

$\angle QPS = 180^\circ - 48^\circ - 48^\circ = 84^\circ$

1. Find $\angle PSR$ using the property of the rhombus.

In rhombus $PQRS$, the diagonal $QS$ bisects the interior angle $\angle PSR$.
$\angle PSR = 2 \times \angle PSQ = 2 \times 48^\circ = 96^\circ$

2. Find $\angle RSV$ using angles on a straight line.

Since $PSV$ is a straight line:
$\angle RSV = 180^\circ - \angle PSR = 180^\circ - 96^\circ = 84^\circ$

3. Find $\angle TSV$ using the property of the parallelogram.

In parallelogram $STUV$, $TU \parallel SV$. Thus, interior angles sum to $180^\circ$.
$\angle TSV = 180^\circ - \angle STU = 180^\circ - 116^\circ = 64^\circ$

4. Calculate $\angle RST$.

$\angle RST = \angle RSV - \angle TSV = 84^\circ - 64^\circ = 20^\circ$

1. Find $\angle QRS$ using the properties of a rhombus.

In rhombus $PQRS$, opposite angles are equal. From part (a), $\angle QPS = 84^\circ$.
$\angle QRS = \angle QPS = 84^\circ$

2. Find $\angle SRT$ using the angle sum of $\triangle RST$.

Using $\angle RST = 20^\circ$ from part (b) and the given $\angle RTS = 21^\circ$:
$\angle SRT = 180^\circ - 20^\circ - 21^\circ = 139^\circ$

3. Calculate $\angle QRT$ using angles at a point.

The angles $\angle QRS$, $\angle SRT$, and $\angle QRT$ meet at point $R$ and complete a full circle ($360^\circ$).
$\angle QRT = 360^\circ - \angle QRS - \angle SRT$
$\angle QRT = 360^\circ - 84^\circ - 139^\circ = 137^\circ$