Question

A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. 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 from the same point is 30°. Find the speed of flying of the bird. 

`("Take"sqrt3=1.732)`

Answer 1

`tan 45° = (AC)/(BC)`

`1 = 80/(BC)`

BC = 80 cm

`tan 30° = (DE)/(BC+CE)`

`1/sqrt3 = 80/(80+x)`

80 + x = 80 √3

x = 80 √3 − 80

x = 80 (√3 − 1) m

The bird travelled 80 (√3 − 1) m

Speed of bird = `"Distance"/"Time"`

`= 40  (sqrt3-1)/2`

= 40 (√3 − 1) m/s

∴ Speed = 40 (√3 − 1) m/s,