1. **State the problem:** Given triangle PQR with point S on line PR such that PSR is straight. 2. Angles PQS and SQR are on a straight line at Q, and their sizes satisfy the ratio $3:1$. 3. Given the angles: $\angle PQS = 69^\circ$ and $\angle SQR = 43^\circ$ from the diagram, but the ratio $3:1$ implies these angles represent parts proportional to 3 and 1 of a total angle at Q. 4. Let $x$ be the measure of the smaller angle $\angle SQR$, then $\angle PQS = 3x$. 5. So, $3x + x = 4x$ represents the angle $\angle PQR$ in triangle PQR. 6. From the diagram, angles $\angle PQS = 69^\circ$ and $\angle SQR = 43^\circ$ given; their sum is $69^\circ + 43^\circ = 112^\circ$. 7. To match the ratio $3:1$, check consistency: $69:43$ simplifies roughly to $1.6:1$, not $3:1$, so instead interpret given ratio as relative parts for different labels. 8. Use ratio to determine exactly $\angle PQS = 3y$, $\angle SQR = y$, so total angle at Q is $4y$. 9. Since PSR is a straight line, $\angle PSQ + \angle SQR = 180^\circ$ (angles on a straight line), or we consider triangle angles summing to $180^\circ$. 10. Using triangle PQR, sum of angles is $69^\circ + 43^\circ + \angle PRQ = 180^\circ$, giving $\angle PRQ = 68^\circ$. 11. Since PSR is a straight line (180°), then $\angle PSQ + \angle SQR = 180^\circ$. 12. $\angle SQR = y = 43^\circ$, so $\angle PSQ = 180^\circ - 43^\circ = 137^\circ$. 13. **Answer:** $\boxed{137^\circ}$ is the size of angle PSQ.