1. **Problem statement:** We are given a figure with parallelogram PQRS and rhombus RSTU sharing side RS. Point S lies on OT. Given angles are $\angle QPS = 66^\circ$ and $\angle RTU = 37^\circ$. We need to find $\angle PSO$. 2. **Key properties:** - In parallelogram PQRS, opposite sides are parallel and equal. - In rhombus RSTU, all sides are equal and opposite angles are equal. - Since S lies on OT, points O, S, T are collinear. 3. **Analyze angles:** - $\angle QPS = 66^\circ$ is given at vertex P. - $\angle RTU = 37^\circ$ is given at vertex T in rhombus RSTU. 4. **Use rhombus properties:** - Rhombus RSTU has equal sides RS = ST = TU = UR. - Opposite angles are equal, so $\angle RST = \angle RTU = 37^\circ$. 5. **Find $\angle PST$ in rhombus:** - Sum of angles in rhombus is 360°. - Adjacent angles are supplementary, so $\angle RST + \angle PST = 180^\circ$. - Thus, $\angle PST = 180^\circ - 37^\circ = 143^\circ$. 6. **Use parallelogram properties:** - In parallelogram PQRS, adjacent angles are supplementary. - $\angle QPS + \angle PSR = 180^\circ$. - Given $\angle QPS = 66^\circ$, so $\angle PSR = 180^\circ - 66^\circ = 114^\circ$. 7. **Relate angles at S:** - $\angle PSR$ and $\angle PST$ share side PS. - Since $\angle PST = 143^\circ$ and $\angle PSR = 114^\circ$, the difference $\angle RST = 143^\circ - 114^\circ = 29^\circ$. 8. **Find $\angle PSO$:** - Since S lies on OT, $\angle PSO + \angle PST = 180^\circ$ (linear pair). - $\angle PST = 143^\circ$, so $\angle PSO = 180^\circ - 143^\circ = 37^\circ$. **Final answer:** $$\boxed{37^\circ}$$