# The angle of elev

In the fig let C be the initial position of the aeroplane. After 15 seconds the position of the aeroplane becomes E.

In ∆ABC

tan 45° =

1 =

Y = 3000m ------(2)

tan 30° =

=

x + y = 3000

Using equation (1) to replace value of y, we get

x = 3000√3 – 3000

3000(√3-1)

2196m

Since the distance travelled by aeroplane in 15 seconds is 2196m. Therefore distance travelled by aeroplane in 1 hour =

527.04 km/hr

Therefore speed of aeroplane is 527.04 km/hr

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