Question

The angle of elevation of an aeroplane from a point on the ground is 45°. After 15 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a height of 3000 m, find the speed of the aeroplane. 

Answer 1

Let A be the point of observation on the ground and B and C be the two positions of aeroplane. Let BL=CM = 3000 m. 

In ΔALB,

`tan 45^circ = "BL"/"AL"`

⇒ `1 = 3000/"AL"`

⇒ `"AL" = 3000`

In ΔAMC,

`tan 30^circ = "MC"/"AM"`

⇒ `1/sqrt(3) = 3000/(3000 + "LM")`

⇒ 3000`sqrt(3)` = (3000 + LM)

⇒ `"LM" = 3000(sqrt(3) - 1)`

∴ BC = `3000(sqrt(3) - 1)`

Now , time taken to travel distance BC = 15 seconds

∴ Speed of the aeroplane = `"Distance"/"Time" = (3000(sqrt(3) - 1))/15 = 200 xx 0.732 = 146.4`

Thus , the speed of the aeroplane is 146.4 m/sec

= `146.4 xx (1/1000)/(1/3600) "km/hr" = 146.4 xx 3.6  "km/hr" = 527.04` km/hr