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प्रश्न
The angle of elevation of an aeroplane from a point on the ground is 45°. After a flight of 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.
उत्तर
Let angle of elevation of an aeroplane is 45°. After 15 seconds angle of elevation is the change to 30°. Let DE be the height of aeroplane which is 3000 meters above the ground.
Let AB = x,BD = y, ∠CAB = 45° and ∠EAD = 30°
Here we have to find speed of aero plane.
We have the corresponding figure as follows
So we use trigonometric ratios.
In Δ ABC
`=> tan A = (BC)/(AB)`
`=> tan 45^@ = 3000/x`
`=> 1 = 3000/x`
`=> x = 3000`
Again in ΔADE
`=> tan A = (DE)/(AB + BD)`
`=> tan 30^@ = 3000/(x + y)`
`=> 1/sqrt3 = 3000/(3000 + y)`
`=> 3000 + y = 3000sqrt3`
`=> y = 3000sqrt3 - 3000`
`=> y = 3000(sqrt3 - 1)`
=> y = 2196
Since 15 sec = 2196
`=> sec = 2196/15 = 146.4`
`= (146.4 xx 3600)/1000`
= 527.04
Hence the speed of aero plane is 527.04 km/h
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