1. **State the problem:** We are given a triangle ΔOPQ with sides $o=290$ inches, $p=720$ inches, and $q=600$ inches. We need to find the measure of angle $\angle P$ to the nearest tenth of a degree. 2. **Recall the Law of Cosines formula:** For any triangle with sides $a$, $b$, and $c$, and angle $C$ opposite side $c$, the Law of Cosines states: $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ 3. **Assign sides and angle:** Here, angle $P$ is opposite side $p=720$. So, $$p^2 = o^2 + q^2 - 2oq\cos(\angle P)$$ 4. **Plug in the known values:** $$720^2 = 290^2 + 600^2 - 2 \times 290 \times 600 \times \cos(\angle P)$$ 5. **Calculate squares:** $$518400 = 84100 + 360000 - 348000 \cos(\angle P)$$ 6. **Simplify the right side:** $$518400 = 444100 - 348000 \cos(\angle P)$$ 7. **Isolate the cosine term:** $$518400 - 444100 = -348000 \cos(\angle P)$$ $$74300 = -348000 \cos(\angle P)$$ 8. **Divide both sides by $-348000$:** $$\cancel{\frac{74300}{-348000}} = \cancel{\frac{-348000 \cos(\angle P)}{-348000}}$$ $$\cos(\angle P) = -0.2138$$ 9. **Find the angle using inverse cosine:** $$\angle P = \cos^{-1}(-0.2138)$$ 10. **Calculate the angle:** $$\angle P \approx 102.3^\circ$$ **Final answer:** $$\boxed{102.3^\circ}$$