Question

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The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 3000√3 m, find the speed of the aeroplane.


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Answer 1

Answer:

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Let P and Q be the two positions of the plane and A be the point of observation. Let ABC be the horizontal line through A. 

It is given that angles of elevation of the plane in two positions P and Q from a point A are 60° and 30° respectively.

∠PAB = 60°, ∠QAB = 30°. It is also given that PB = 3000√3 meters

In Δ ABP, we have

`tan 60="BP"/"AB"`

`s q r t 3/1=(3000sqrt3)/"AB"`

AB=3000m

`tan 30="QC"/"AC"`

AC = 9000 m

∴ Distance = BC = AC – AB = 9000m – 3000m = 6000m

Thus, the plane travels 6km in 30 seconds

Hence speed of plane = 6000/30 = 200 m/sec = 720km/h.

Answer 2

$\sf\huge\bold\pink{Answer:}$

Let P and Q be the two positions of the plane and A be the point of observation. Let ABC be the horizontal line through A. 

It is given that angles of elevation of the plane in two positions P and Q from a point A are 60° and 30° respectively.

∠PAB = 60°, ∠QAB = 30°. It is also given that PB = 3000√3 meters

In Δ ABP, we have

`tan 60="BP"/"AB"`

`s q r t 3/1=(3000sqrt3)/"AB"`

AB=3000m

`tan 30="QC"/"AC"`

AC = 9000 m

∴ Distance = BC = AC – AB = 9000m – 3000m = 6000m

Thus, the plane travels 6km in 30 seconds

Hence speed of plane = 6000/30 = 200 m/sec = 720km/h.