1. **State the problem:** We need to find the length of side $QR$ in quadrilateral $PQRS$, which is formed by two triangles $PQS$ and $QRS$. 2. **Given data:** - Triangle $PQS$: $PQ=21m$, $QS=18m$, angle $PSQ=125^\circ$, angle $QPS=25^\circ$. - Triangle $QRS$: $QS=18m$, $SR=16m$, angle $QSR=125^\circ$ (since $\angle PSQ=125^\circ$ and $S$ is common vertex), angle $QRS$ unknown. 3. **Goal:** Calculate length $QR$. 4. **Approach:** Use the Law of Cosines in triangle $QRS$ to find $QR$. 5. **Law of Cosines formula:** $$c^2 = a^2 + b^2 - 2ab \cos(C)$$ where $c$ is the side opposite angle $C$. 6. **Identify sides and angle in triangle $QRS$:** - $a = QS = 18$ - $b = SR = 16$ - $C = \angle QSR = 125^\circ$ - $c = QR$ (unknown) 7. **Apply Law of Cosines:** $$QR^2 = 18^2 + 16^2 - 2 \times 18 \times 16 \times \cos(125^\circ)$$ 8. **Calculate values:** $$18^2 = 324$$ $$16^2 = 256$$ $$\cos(125^\circ) = \cos(180^\circ - 55^\circ) = -\cos(55^\circ) \approx -0.5736$$ 9. **Substitute:** $$QR^2 = 324 + 256 - 2 \times 18 \times 16 \times (-0.5736)$$ $$QR^2 = 580 + 2 \times 18 \times 16 \times 0.5736$$ 10. **Calculate product:** $$2 \times 18 \times 16 = 576$$ $$576 \times 0.5736 \approx 330.45$$ 11. **Sum:** $$QR^2 = 580 + 330.45 = 910.45$$ 12. **Find $QR$:** $$QR = \sqrt{910.45} \approx 30.17m$$ **Final answer:** $$\boxed{QR \approx 30.17\text{ meters}}$$