1. **State the problem:** We need to find the angle between two lines of sight from a point 20 meters away from the Ferris wheel's center to a rider at times $t=45$ seconds and $t=57$ seconds. 2. **Identify given data:** - Diameter of Ferris wheel: 16 m, so radius $r=8$ m. - Lowest point is 2 m above ground, so center height $= 2 + 8 = 10$ m. - One revolution every 30 seconds, so angular speed $\omega = \frac{2\pi}{30} = \frac{\pi}{15}$ radians per second. - Distance from observer to center horizontally: 20 m. 3. **Find the angular position of the rider at each time:** - Angle at time $t$ is $\theta(t) = \omega t = \frac{\pi}{15} t$. - At $t=45$ s: $\theta_1 = \frac{\pi}{15} \times 45 = 3\pi$ radians. - At $t=57$ s: $\theta_2 = \frac{\pi}{15} \times 57 = \frac{57\pi}{15} = 3.8\pi$ radians. 4. **Simplify angles modulo $2\pi$ (since angles repeat every revolution):** - $3\pi = 3\pi - 2\pi = \pi$ radians. - $3.8\pi = 3.8\pi - 2\pi = 1.8\pi$ radians. 5. **Calculate the vertical position of the rider at each time:** - Vertical height $y(t) = 10 + 8 \sin(\theta(t))$. - At $\theta_1 = \pi$: $y_1 = 10 + 8 \sin(\pi) = 10 + 0 = 10$ m. - At $\theta_2 = 1.8\pi$: $y_2 = 10 + 8 \sin(1.8\pi) = 10 + 8 \sin(\pi + 0.8\pi) = 10 - 8 \sin(0.8\pi)$. - Calculate $\sin(0.8\pi) = \sin(144^\circ) \approx 0.5878$. - So $y_2 = 10 - 8 \times 0.5878 = 10 - 4.7024 = 5.2976$ m. 6. **Calculate horizontal distance from observer to rider (constant):** - Horizontal distance $x = 20$ m. 7. **Calculate angles of lines of sight from observer to rider at each time:** - Angle $\alpha = \arctan\left(\frac{y - 0}{x}\right) = \arctan\left(\frac{y}{20}\right)$. - At $t=45$ s: $\alpha_1 = \arctan\left(\frac{10}{20}\right) = \arctan(0.5) \approx 0.464$ radians. - At $t=57$ s: $\alpha_2 = \arctan\left(\frac{5.2976}{20}\right) = \arctan(0.2649) \approx 0.259$ radians. 8. **Find the angle between the two lines of sight:** - $\Delta \alpha = |\alpha_1 - \alpha_2| = |0.464 - 0.259| = 0.205$ radians. - Convert to degrees: $0.205 \times \frac{180}{\pi} \approx 11.75^\circ$. 9. **Interpretation for safety:** - The angle between sightlines is about $11.75^\circ$, which helps ensure that the rider's position changes visibly and that the Ferris wheel's structure and nearby obstacles can be monitored for clearance. - Knowing these angles helps in planning safe distances and sightlines to avoid collisions or obstructions. **Final answer:** The approximate angle between the two lines of sight is $\boxed{0.205}$ radians or about $\boxed{11.75^\circ}$.