Question

A bird is sitting on the top of a 80m gh tree. from a point on the ground, the angle of elevation of the bird is 45⁰. the bird flies away horizontally in such a way that it remained at a constant height from the ground. after 2 seconds, the angle of elevation of the bird form the same point is 30⁰. find the speed of flying of the bird

Answer 1

Let the speed of the birs be x m/s. 
So, the distance covered by it in 2 seconds will be 2x ( distance = speed x time)
When on top of the tree, the angle of elevation of the bird from point O (see figure) is 45.
tan 45 = AB/BO = 1
So, AB=BO=80m.
After flying for 2 seconds, angle of elevation from O is 30.
tan 30 = CD/DO= 1/√3
DO= 80√3m

distance covered by bird in 2 seconds = BO-DO = 80 -80√3
2x= 80(1-√3)
Therefore, x = 40(1-√3) m/s