1. Given a triangle PQR with sides $PQ = 12$ cm, $QR = 9$ cm, and $PR = 18$ cm, we need to find the angle $\theta$ at vertex $Q$ between sides $PQ$ and $QR$. 2. We can use the Law of Cosines which states: $$PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos \theta$$ 3. Substitute the known values: $$18^2 = 12^2 + 9^2 - 2 \times 12 \times 9 \times \cos \theta$$ 4. Simplify each term: $$324 = 144 + 81 - 216 \cos \theta$$ $$324 = 225 - 216 \cos \theta$$ 5. Rearrange to isolate $\cos \theta$: $$216 \cos \theta = 225 - 324$$ $$216 \cos \theta = -99$$ 6. Divide both sides by 216: $$\cos \theta = \frac{-99}{216} = -\frac{11}{24}$$ 7. Calculate $\theta$ by taking the inverse cosine (in degrees): $$\theta = \cos^{-1} \left(-\frac{11}{24}\right) \approx 117.3^\circ$$ **Final answer:** $$\theta \approx 117.3^\circ$$