1. **Stating the problem:** We have a composite figure with parallelogram PQRS and rhombus RSTU. Given $\angle QPS = 66^\circ$ and $\angle RTU = 37^\circ$, we need to find $\angle PSO$. 2. **Key properties:** - In parallelogram PQRS, opposite sides are parallel and equal, and opposite angles are equal. Adjacent angles are supplementary. - In rhombus RSTU, all sides are equal, and opposite angles are equal. Adjacent angles are supplementary. - Point S lies on line segment OT, so $\angle PSO$ is formed at S between points P and O. 3. **Analyze parallelogram PQRS:** Given $\angle QPS = 66^\circ$, this is an angle at vertex P between points Q and S. Since PQRS is a parallelogram, $\angle QPS = \angle PSR = 66^\circ$ because opposite angles are equal. 4. **Analyze rhombus RSTU:** Given $\angle RTU = 37^\circ$, this is an angle at vertex T between points R and U. Since RSTU is a rhombus, adjacent angles are supplementary, so $\angle STR = 180^\circ - 37^\circ = 143^\circ$. 5. **Find $\angle PSO$:** Since S lies on OT, and OT is a line segment, $\angle PSO$ is the angle at S between points P and O. Because $\angle PSR = 66^\circ$ and $\angle STR = 143^\circ$, and points R, S, T, O are collinear or connected, $\angle PSO = 180^\circ - 66^\circ - 37^\circ = 77^\circ$. **Final answer:** $$\boxed{77^\circ}$$