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Question
An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.
Solution
An aeroplane is flying 1 km above the ground making an angle of elevation of aeroplane 60°. After 10 seconds angle of elevation is changed to 30°.
Let CE = x, BC = y, ∠AEB = 30°, ∠DEC = 60°, AB = 1 km and CD = 1 km.
Here we have to find the speed of an aeroplane.
The corresponding figure is as follows
So we use trigonometric ratios.
In ΔDCE
`=> tan 60^@ = 1/x`
`=> sqrt3 = 1/x`
`=> x = 1/sqrt3`
Again in Δ ABE
`=> tan 30^@ = 1/(x + y)`
`=> 1/sqrt3 = 1/(x + y)`
`=> x + y = sqrt3`
`=> y = sqrt3 - 1/sqrt3`
`=> y = 2/sqrt3``
speed = "distance"/"time"`
`= y/(10 sec)`
`= (2/sqrt3)/(10/(60 xx 60))`
= 415.68
Hence the speed of aeroplace is 415.68 km/h.
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