Asked by kelvin
A plane leaves an airport x,20-60E and 36-80N and flies due south along the same longitude for 8hours at the rate of 1000km/h to another airport Y,20-60E and titaS.the plane then flies west to another airport Z for 8hours at the same speed. Calculate to the nearest degree (a) the value of tita. (b) the longitude of Z
Answers
Answered by
Steve
wow - can you make heads or tails of that gibberish? I assume that 22-60 means 22°60'. No idea what "tita" means. Must be "theta".
At any rate, since the circumference of the earth is about 40,000 km, the plane flew 8000km = 1/5 of the way around the earth. That is 360/5 = 72°. Starting from 36°80'N, that puts him at Y=35°20'S.
The length of a line of latitude at an angle x is 40000 cos(x) km. That means that the "circumference" of the earth at 35°20'S is 40000*0.8158 = 32632km.
At that latitude, 8000km is 0.245*360° = 88°15' westward. Starting from 20°60'E, he ends up at longitude 67°15'W.
Hmmm. I just noticed that 22-60E cannot be 22°60', since 60' = 1° so no one would write that. So, fix my misinterpretation and follow the steps to the correct result.
At any rate, since the circumference of the earth is about 40,000 km, the plane flew 8000km = 1/5 of the way around the earth. That is 360/5 = 72°. Starting from 36°80'N, that puts him at Y=35°20'S.
The length of a line of latitude at an angle x is 40000 cos(x) km. That means that the "circumference" of the earth at 35°20'S is 40000*0.8158 = 32632km.
At that latitude, 8000km is 0.245*360° = 88°15' westward. Starting from 20°60'E, he ends up at longitude 67°15'W.
Hmmm. I just noticed that 22-60E cannot be 22°60', since 60' = 1° so no one would write that. So, fix my misinterpretation and follow the steps to the correct result.
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