a bird sitting on the top of a tree, which is 80m high. the angle of elevation of the bird, from the point on the ground is 45^{0}. the bird flies away from the point of observation horizantally and remains at a constant height. after 2 seconds, the angle of elevation of the bird from the point of observation becomes 30^{0}. find the sped of the flying bird

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

Given: The angle of elevations of the bird in two positions P and Q from point A are 45° and 30°.

∴ ∠PAB = 45°, ∠QAB = 30°.

Also,

PB = 80 meters

In ΔABP, we have,

In ΔACQ, we have,

∴ PQ = BC = AC – AB =

Thus,

The bird touch in 2 seconds.

Hence,

Speed of the bird

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