1. **Problem statement:** We are given a triangle PQR with sides $PQ=15$, $PR=9$, and the included angle $\angle P=136^\circ$. We need to find the length $p$ of the side opposite angle $P$. 2. **Formula used:** To find the side opposite a known angle in a triangle when two sides and the included angle are known, we use the Law of Cosines: $$p^2 = PQ^2 + PR^2 - 2 \times PQ \times PR \times \cos(\angle P)$$ 3. **Substitute the known values:** $$p^2 = 15^2 + 9^2 - 2 \times 15 \times 9 \times \cos(136^\circ)$$ 4. **Calculate each term:** $$15^2 = 225$$ $$9^2 = 81$$ $$2 \times 15 \times 9 = 270$$ 5. **Calculate $\cos(136^\circ)$:** Since $136^\circ$ is in the second quadrant, $\cos(136^\circ)$ is negative. Using a calculator: $$\cos(136^\circ) \approx -0.7193$$ 6. **Plug in the cosine value:** $$p^2 = 225 + 81 - 270 \times (-0.7193)$$ $$p^2 = 306 + 194.211$$ $$p^2 = 500.211$$ 7. **Find $p$ by taking the square root:** $$p = \sqrt{500.211} \approx 22.4$$ **Final answer:** $$p \approx 22.4$$ The length of side $p$ is approximately 22.4 units.