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.
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°
PB = 80 m
In ΔABP, we have
⇒ AB=80 m
In ΔACQ, we have
∴ 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