1. **State the problem:** We have triangle $\triangle OPQ$ with side $p = 9.7$ inches, angle $\angle Q = 93^\circ$, and angle $\angle O = 82^\circ$. We need to find the length of side $q$ to the nearest tenth of an inch. 2. **Find the missing angle:** The sum of angles in a triangle is $180^\circ$. $$\angle P = 180^\circ - 93^\circ - 82^\circ = 5^\circ$$ 3. **Use the Law of Sines:** The Law of Sines states: $$\frac{p}{\sin \angle P} = \frac{q}{\sin \angle Q}$$ 4. **Plug in known values:** $$\frac{9.7}{\sin 5^\circ} = \frac{q}{\sin 93^\circ}$$ 5. **Solve for $q$:** $$q = \frac{9.7 \times \sin 93^\circ}{\sin 5^\circ}$$ 6. **Calculate sine values:** $$\sin 93^\circ \approx 0.9986, \quad \sin 5^\circ \approx 0.0872$$ 7. **Evaluate $q$:** $$q = \frac{9.7 \times 0.9986}{0.0872} = \frac{9.686}{0.0872} \approx 111.1$$ 8. **Final answer:** The length of side $q$ is approximately $111.1$ inches to the nearest tenth.