Answer :

Let D be the initial position of the aeroplane, 2500 m above the ground. After 15 s, let A be the position of the aeroplane flying in the same height. Draw two lines AB and DE from A and D perpendicular to the ground. Then, AB = DE = 2500m. Let C be the point on the ground from which the angles of elevation of the two positions of the plane, viz., D and A are 45° and 30° respectively. Join A and D with C. Also join the points B,E and C on the ground. We then get two right-angled triangles ABC and DEC with right angles at B and E respectively. So, we get ∠ACB = 30° and ∠DCE = 45°. We are to find speed of the plane. The plane travels a distance AD in 15 s. Since speed = (distance/time), we need to find AD only.

We first find CE from ∆DEC using the trigonometric ratio tan θ.

In ∆DEC,

or,

or,

or,

Now, we will again use tan to find the value of BE from ∆ABC.

In ∆ABC,

or,

or,

or,

Now, BE = AD = 1830 m.

Speed of the aeroplane = .

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