Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10th

# 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 consta - Mathematics

Sum

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)

#### Solution

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

Is there an error in this question or solution?

#### APPEARS IN

Samacheer Kalvi Mathematics Class 10 SSLC Tamil Nadu State Board
Chapter 6 Trigonometry
Unit Exercise – 6 | Q 5 | Page 267