An aeroplane is flying at a height of 3000 m above the ground level. If the angle subtended at ground observation point by the aeroplane position 12 s apart is 30°, what is the speed of the aeroplane?
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Answer:
right angled triangle OAB, tan30
= h/x
We get x= h/tan30°
= 3000√3
=5190
Time taken t=12
Thus speed of the aircraft v= x/t
=5190/12
= 432.5 m/s
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