1. **State the problem:** We need to find the smallest angle in triangle PQR with sides 9 cm, 10 cm, and 14 cm. 2. **Identify the smallest angle:** The smallest angle is opposite the shortest side. The sides are 9 cm, 10 cm, and 14 cm, so the smallest angle is opposite the side of length 9 cm. 3. **Use the cosine rule:** The cosine rule states: $$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$ where $A$ is the angle opposite side $a$, and $b$ and $c$ are the other two sides. 4. **Assign sides:** Let $a = 9$ cm (side opposite the smallest angle), $b = 10$ cm, and $c = 14$ cm. 5. **Calculate cosine of the smallest angle:** $$\cos A = \frac{10^2 + 14^2 - 9^2}{2 \times 10 \times 14} = \frac{100 + 196 - 81}{280} = \frac{215}{280} = 0.7679$$ 6. **Find the angle:** $$A = \cos^{-1}(0.7679) \approx 40.1^\circ$$ 7. **Final answer:** The smallest angle in triangle PQR is approximately $40.1^\circ$ (to 3 significant figures).