1. **Problem statement:** We are given a parallelogram $PQRS$ with side $RS=5$, side $SP=3$, and angle $Q=128^\circ$. We need to find the length of side $QP$. 2. **Recall properties of parallelograms:** Opposite sides are equal and parallel. Thus, $PQ = SR = 5$ and $SP = QR = 3$. The angle at $Q$ is given as $128^\circ$. 3. **Use the Law of Cosines:** To find $QP$, consider triangle $PQR$. Since $PQRS$ is a parallelogram, $QP$ is the side opposite angle $R$ or $Q$. We can use the Law of Cosines on triangle $PQR$ with sides $PQ$, $QR$, and angle $Q$. 4. **Apply the Law of Cosines formula:** $$QP^2 = SP^2 + PQ^2 - 2 \times SP \times PQ \times \cos(128^\circ)$$ 5. **Substitute known values:** $$QP^2 = 3^2 + 5^2 - 2 \times 3 \times 5 \times \cos(128^\circ)$$ 6. **Calculate:** $$QP^2 = 9 + 25 - 30 \times \cos(128^\circ) = 34 - 30 \times \cos(128^\circ)$$ 7. **Evaluate $\cos(128^\circ)$:** $$\cos(128^\circ) \approx -0.6157$$ 8. **Plug in the cosine value:** $$QP^2 = 34 - 30 \times (-0.6157) = 34 + 18.471 = 52.471$$ 9. **Find $QP$ by taking the square root:** $$QP = \sqrt{52.471} \approx 7.24$$ **Final answer:** $$\boxed{QP \approx 7.24}$$