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 450. 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 300. 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°.


PB = 80 meters


In ΔABP, we have,


In ΔACQ, we have,


∴ PQ = BC = AC – AB =



The bird touch  in 2 seconds.



Speed of the bird


  • 206

the speed is 40(root3-1) ms-1

  • -11

aha...! great but can i have the soluion with calculation...... since i am really weak in maths

  • -41
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