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An aeroplane flying horizontally 1 km above ground is observed at an elevation of 60° . After 10 seconds elevation is observed to be 30°. Find speed in



When the elevation was 60, distance of the point of observation to the point on the groundbelow the plane = 1xcot60 = 1/?3 km Similarly when the angle becomes 30, distance becomes = 1xcot30 = ?3 km So, distance travelled = ?3 -1/?3 km = 2/?3 km time taken = 10sec = 10/3600 hr hence speed = distance / time = 2x3600/10x?3 = 720/?3 km/hr = 415.7 km/hr
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and it was supposed to be tan instead of cot i think
No, he used cot values instead of tan
Idk why, but ok
i just solved using tan got correct answer lol
Yeah it doesn't really make any difference, I find it easier to use tan
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