The plane took off from a point with coordinates 50 * N. 0 v.d. and flew 5550 km south. Then he turned west and flew 2,220 km. Turning again, he flew another 1110 km to the north. At which point on the map he ended up.
To solve the problem, you need to express these distances in degrees. It is known that the length of the 1 * arc of any meridian is 111 km. First, the plane flew south. He flew the distance (5550: 111 = 50) 50 *. Thus, it turns out that, having taken off from a point with coordinates 50 * s. sh. and 0 * v.d. and flying 50 * south along the prime meridian, he found himself at the equator at the intersection of the equator and the meridian. Then the plane flew west along the equator. The length of the 1 * arc of the parallel is different at different latitudes, but the length of the 1 * arc of the equator is known – 111 km. Therefore, you can determine the distance that the plane flew along the equator: (2220: 111 = 20) 20 * Further, the plane flew from the equator along the meridian 20 * W. d. to the north. He flew a distance of 1110 km, i.e. 10 * (1110: 111 = 10) Accordingly, he was at a point with coordinates 10 * N. and 20 * W
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