1. **Problem statement:** We are given a triangle QRP with points R, S, and P on a straight line (RSP). The angle $\angle QSP = 80^\circ$ and line QS is perpendicular to RS. We need to find the measure of $\angle PQR$. 2. **Understanding the problem:** Since R, S, and P are collinear and $\angle QSP = 80^\circ$, and QS is perpendicular to RS, $\angle QSR = 90^\circ$. 3. **Key facts:** - $\angle QSP + \angle RSP = 180^\circ$ because RSP is a straight line. - $\angle QSR = 90^\circ$ because QS is perpendicular to RS. - Triangle QSP is isosceles with QS = QS (given), so $\angle PQS = \angle PQR$. 4. **Calculate $\angle RSP$:** $$\angle RSP = 180^\circ - 80^\circ = 100^\circ$$ 5. **Calculate $\angle QSR$:** Given QS is perpendicular to RS, so $$\angle QSR = 90^\circ$$ 6. **Find $\angle PSQ$:** Since $\angle QSP = 80^\circ$, and $\angle QSR = 90^\circ$, the triangle QSP has angles at S summing to 180°: $$\angle PSQ = 180^\circ - 80^\circ - 90^\circ = 10^\circ$$ 7. **Use triangle QSP to find $\angle PQR$:** Since QS is perpendicular to RS and triangle QSP is isosceles with QS = QS, $\angle PQR = \angle PQS = 10^\circ$. **Final answer:** $$\boxed{10^\circ}$$