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 away 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°. Determine the speed at which the bird flies `(sqrt(3) = 1.732)`

Answer 1

A is the initial position of the bird B is the final position of the bird Let the speed of the bird be s

Distance = speed × time

∴ AB = 2x

Let CD be x

∴ CE = x + 2s

In the ∆CDA, tan 45° = `"AD"/"CD"`

1 = `80/x`

x = 80   ...(1)

In the ∆BCE

tan 30° = `"BE"/"CE"`

`1/sqrt(3) = 80/(x + 2"s")`

x + 2s = `80sqrt(3)`

x = `80sqrt(3) - 2"s"`   ...(2)

From (1) and (2) we get

`80 sqrt(3) - 2"s"` = 80

`80 sqrt(3) - 80` = 2s

⇒ `80(sqrt(3) - 1)` = 2s

s = `(80(sqrt(3) - 1))/2`

= 40(1.732 – 1)

= 40 × 0.732

= 29.28

Speed of the flying bird = 29.28 m/sec