1. **State the problem:** We are given a triangle PQR with sides $|PQ|=15$ cm, $|PR|=12$ cm, and $|RQ|=5$ cm. We need to find the measure of angle $\angle PQR$ to the nearest degree. 2. **Identify the angle and sides:** $\angle PQR$ is the angle at vertex Q. The sides adjacent to Q are $|PQ|=15$ cm and $|RQ|=5$ cm, and the side opposite to $\angle PQR$ is $|PR|=12$ cm. 3. **Use the Law of Cosines:** The Law of Cosines relates the sides and angles of a triangle: $$c^2 = a^2 + b^2 - 2ab \cos C$$ where $C$ is the angle opposite side $c$, and $a$ and $b$ are the other two sides. 4. **Apply the formula:** Here, $C = \angle PQR$, $c = |PR| = 12$, $a = |PQ| = 15$, and $b = |RQ| = 5$. $$12^2 = 15^2 + 5^2 - 2 \times 15 \times 5 \times \cos \angle PQR$$ 5. **Calculate squares:** $$144 = 225 + 25 - 150 \cos \angle PQR$$ 6. **Simplify:** $$144 = 250 - 150 \cos \angle PQR$$ 7. **Isolate cosine term:** $$150 \cos \angle PQR = 250 - 144$$ $$150 \cos \angle PQR = 106$$ 8. **Divide both sides:** $$\cos \angle PQR = \frac{106}{150}$$ $$\cos \angle PQR = \frac{\cancel{106}}{\cancel{150}}$$ (no common factors to cancel, so fraction remains as is) 9. **Calculate the cosine value:** $$\cos \angle PQR = 0.7067$$ (rounded to 4 decimal places) 10. **Find the angle:** Use the inverse cosine function: $$\angle PQR = \cos^{-1}(0.7067)$$ 11. **Calculate the angle:** $$\angle PQR \approx 45.0^\circ$$ 12. **Final answer:** The measure of $\angle PQR$ is approximately **45 degrees** to the nearest degree.