1. **Problem statement:** Given a circle with center O, chord PQ of length 8 cm, N is the midpoint of PQ, ON = 3 cm, and \(\angle ONR = 20^\circ\). We need to find the size of \(\angle ORN\) to the nearest degree. 2. **Understand the setup:** Since N is the midpoint of chord PQ, \(PN = NQ = \frac{8}{2} = 4\) cm. 3. **Right triangle ONP:** Because ON is perpendicular to PQ at N, triangle ONP is right angled at N. 4. **Calculate OP:** Using Pythagoras theorem in triangle ONP, $$OP = \sqrt{ON^2 + PN^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm}.$$ 5. **Radius of the circle:** OP is the radius, so radius \(r = 5\) cm. 6. **Analyze triangle ONR:** We know \(\angle ONR = 20^\circ\), ON = 3 cm, and OR = radius = 5 cm. 7. **Find \(\angle ORN\) using Law of Cosines or Law of Sines:** Use Law of Sines in triangle ONR: $$\frac{\sin(\angle ORN)}{ON} = \frac{\sin(\angle ONR)}{OR}$$ Let \(x = \angle ORN\), then $$\frac{\sin x}{3} = \frac{\sin 20^\circ}{5}$$ 8. **Calculate \(\sin x\):** $$\sin x = 3 \times \frac{\sin 20^\circ}{5} = \frac{3}{5} \times \sin 20^\circ$$ Calculate \(\sin 20^\circ \approx 0.3420\), so $$\sin x = \frac{3}{5} \times 0.3420 = 0.2052$$ 9. **Find \(x\):** $$x = \sin^{-1}(0.2052) \approx 11.8^\circ$$ 10. **Final answer:** The size of \(\angle ORN\) is approximately \(12^\circ\) to the nearest degree.