1 Multiple Choice 1 point Assume that the cities lie on the same north-south line and that the radius of the earth is 6400 km . Find the distance between City \( \mathrm{E}, 20^{\circ} \mathrm{N} \) and City F, \( 58^{\circ} \mathrm{S} \). (Round to the nearest kilometer.)

1. The two cities are located on the same meridian. The angular difference between City \( \mathrm{E} \) at \( 20^\circ\mathrm{N} \) and City \( \mathrm{F} \) at \( 58^\circ\mathrm{S} \) is \[ 20^\circ + 58^\circ = 78^\circ. \] 2. Convert \( 78^\circ \) to radians (using the conversion \( \pi \) radians = \( 180^\circ \)): \[ 78^\circ = 78 \times \frac{\pi}{180} = \frac{78\pi}{180}. \] 3. The distance along the Earth’s surface (the arc length) is given by \[ \text{Distance} = R \times \theta, \] where \( R = 6400 \) km and \( \theta \) is in radians. Therefore, the distance is \[ 6400 \times \frac{78\pi}{180}. \] 4. Simplify the expression: \[ \frac{78}{180} = \frac{13}{30}. \] Thus, the distance becomes \[ 6400 \times \frac{13\pi}{30}. \] 5. Approximating using \( \pi \approx 3.1416 \): \[ \text{Distance} \approx 6400 \times \frac{13 \times 3.1416}{30} \approx 6400 \times \frac{40.8408}{30} \approx 6400 \times 1.36136 \approx 8713 \text{ km}. \] Therefore, the distance between City \( \mathrm{E} \) and City \( \mathrm{F} \) is approximately \( 8713 \) kilometers.