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)`

#### Solution

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 angles of elevations of the bird in two positions P and Q from point A are 45° and 30°, respectively.

∴ ∠PAB = 45° and ∠QAB = 30°

Also,

PB = 80 m

In ΔABP, we have

`tan45^@=(BP)/(AB)`

`=>1=80/(AB)`

⇒ AB=80 m

In ΔACQ, we have

`tan30^@=(CQ)/(AC)`

`=>1/sqrt3=80/(AC)`

`=>AC=800sqrt3 m`

∴ PQ = BC = AC – AB = 803√−80=80 (`sqrt3`−1) m

So, the bird covers 80 (`sqrt3`- 1)min 2 s

Thus, speed of the bird is given by

`=(80(sqrt3-1))/2`

`=(40(sqrt3-1)xx60xx60)/1000`

=144(1.732−1)km/h

=105.408 km/h