1. **State the problem:** We need to find the length of side $OP$ in rectangle $OPRS$ given angles and side lengths. 2. **Given:** $\angle R = 90^\circ$, $\angle S = 90^\circ$, $\angle RQO = 39.7^\circ$, $\angle SQP = 50.3^\circ$, $RQ = 50$ m, $QS = 40$ m. 3. **Understand the figure:** $OPRS$ is a rectangle, so $OR$ and $PS$ are parallel and equal, and $RS$ and $OP$ are parallel and equal. 4. **Use trigonometry in triangles involving $Q$:** Points $Q$ lies at the intersection of diagonals $OQ$ and $PQ$. 5. **Calculate $OQ$ using triangle $RQO$:** - In right triangle $RQO$, $\angle RQO = 39.7^\circ$, side $RQ = 50$ m. - Use cosine to find $OQ$: $$\cos(39.7^\circ) = \frac{OQ}{RQ} \implies OQ = RQ \times \cos(39.7^\circ) = 50 \times \cos(39.7^\circ)$$ 6. **Calculate $PQ$ using triangle $SQP$:** - In right triangle $SQP$, $\angle SQP = 50.3^\circ$, side $QS = 40$ m. - Use cosine to find $PQ$: $$\cos(50.3^\circ) = \frac{PQ}{QS} \implies PQ = QS \times \cos(50.3^\circ) = 40 \times \cos(50.3^\circ)$$ 7. **Since $OQ$ and $PQ$ are parts of the diagonals intersecting at $Q$, and $OP$ is the diagonal of the rectangle, the length $OP = OQ + PQ$** 8. **Calculate values:** $$OQ = 50 \times \cos(39.7^\circ) \approx 50 \times 0.768 = 38.4$$ $$PQ = 40 \times \cos(50.3^\circ) \approx 40 \times 0.639 = 25.6$$ 9. **Sum to find $OP$:** $$OP = OQ + PQ = 38.4 + 25.6 = 64.0$$ 10. **Final answer:** The length of $OP$ is approximately **64.0 metres** to the nearest tenth.