1. **State the problem:** We need to find the missing angle $P$ in a triangle where one angle is $20^\circ$ and the sides adjacent to angle $P$ are 15 cm and 18 cm, with the side opposite the $20^\circ$ angle being 29 cm. 2. **Identify the known elements:** - Side opposite $20^\circ$ is 29 cm. - Sides adjacent to angle $P$ are 15 cm and 18 cm. 3. **Use the Law of Cosines to find angle $P$:** The Law of Cosines states: $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ where $C$ is the angle opposite side $c$. 4. **Assign sides:** Let $a = 15$, $b = 18$, and $c = 29$ (opposite the $20^\circ$ angle). We want to find angle $P$ between sides 15 and 18. 5. **Calculate the length of the side opposite angle $P$ (which is 31 cm as per the image):** Since the triangle has sides 15, 18, and 31 cm, and angle $P$ is between sides 15 and 18, side opposite $P$ is 31 cm. 6. **Apply Law of Cosines to find angle $P$:** $$31^2 = 15^2 + 18^2 - 2 \times 15 \times 18 \times \cos(P)$$ $$961 = 225 + 324 - 540 \cos(P)$$ $$961 = 549 - 540 \cos(P)$$ 7. **Isolate $\cos(P)$:** $$961 - 549 = -540 \cos(P)$$ $$412 = -540 \cos(P)$$ $$\cos(P) = \frac{\cancel{412}}{\cancel{-540}} = -\frac{412}{540}$$ 8. **Calculate $\cos(P)$:** $$\cos(P) = -0.76296$$ 9. **Find angle $P$ using inverse cosine:** $$P = \cos^{-1}(-0.76296)$$ $$P \approx 139.7^\circ$$ 10. **Round to the nearest integer:** $$P \approx 140^\circ$$ **Final answer:** The missing angle $P$ is approximately $140^\circ$.