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Question
The angle of elevation of an aeroplane from point A on the ground is 60˚. After flight of 15 seconds, the angle of elevation changes to 30˚. If the aeroplane is flying at a constant height of 1500√3 m, find the speed of the plane in km/hr.
Solution
Let BC be the height at which the aeroplane is observed from point A.
Then, BC = 1500√3
In 15 seconds, the aeroplane moves from point A to D.
A and D are the points where the angles of elevation 60 and 30
are formed respectively.
Let BA = x metres and AD y metres
BC = x + y
In ΔCBA,
`tan60^@="BC"/"BA"`
`sqrt3=(1500sqrt3)/x`
x=1500 m........(1)
In ΔCBD,
`tan30^@="BC"/"BD"`
`1/sqrt3=(1500sqrt3)/(x+y)`
x + y =1500(3)= 4500
1500 + y = 4500
y = 3000 m ....(2)
We know that the aeroplane moves from point A to D in 15 seconds and the distance covered is 3000 metres. (by 2)
`"speed"="distance"/"time"`
`"speed"=3000/15`
speed=200 m/s
Converting it to km/hr =200x(18/5)=720 km/hr
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