Answer :

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)


In ∆ADE


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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