1. **Problem statement:** We are given the offset distance from PC (Point of Curvature) to PT (Point of Tangency) of a simple curve as 8 meters, and the length of the curve as 60 meters. We need to find the radius $R$ of the curve. 2. **Relevant formula:** For a simple circular curve, the length $L$, radius $R$, and offset distance $E$ (also called the middle ordinate) are related by the formula: $$E = R \left(1 - \cos\left(\frac{L}{2R}\right)\right)$$ 3. **Explanation:** - $E$ is the offset distance from the midpoint of the chord to the arc. - $L$ is the length of the curve (arc length). - $R$ is the radius of the curve. 4. **Given values:** $$E = 8, \quad L = 60$$ 5. **Substitute values into the formula:** $$8 = R \left(1 - \cos\left(\frac{60}{2R}\right)\right) = R \left(1 - \cos\left(\frac{30}{R}\right)\right)$$ 6. **Rearranged equation:** $$\frac{8}{R} = 1 - \cos\left(\frac{30}{R}\right)$$ 7. **Solve for $R$ numerically:** This transcendental equation cannot be solved algebraically, so we use numerical methods (trial and error or a calculator). Try $R=50$: $$\frac{8}{50} = 0.16$$ $$1 - \cos\left(\frac{30}{50}\right) = 1 - \cos(0.6) \approx 1 - 0.8253 = 0.1747$$ Since $0.16 < 0.1747$, try a slightly larger $R$. Try $R=55$: $$\frac{8}{55} \approx 0.1455$$ $$1 - \cos\left(\frac{30}{55}\right) = 1 - \cos(0.5455) \approx 1 - 0.8555 = 0.1445$$ Close, but $0.1455 > 0.1445$, so $R$ is near 55. Try $R=54.5$: $$\frac{8}{54.5} \approx 0.1468$$ $$1 - \cos\left(\frac{30}{54.5}\right) = 1 - \cos(0.5505) \approx 1 - 0.8527 = 0.1473$$ Very close. 8. **Final answer:** $$R \approx 54.5 \text{ meters}$$ This is the radius of the curve.