Question

A bird is sitting on the top of a 60 m high tree. From a point on the ground, the angle of
elevation of the bird is 60 degree. 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 degree. Find the speed of the flying bird.

Answer 1

Answer:

20√3 m/s

Step-by-step explanation:

tana=p/b

p=height of tree=60m

tan60°=60/b

√3=60/b

b=60/√3

b=60*√3 / √3*√3

b=60√3 / 3

b=20√3

tanb=p/b

tan30°=60/b

1/√3 = 60/b

b=60√3m

distance travelled=60√3 - 20√3

=√3(60-20)

=40√3m

speed =distance/time

=40√3/2

= 20√3 m/s

Answer 2

★ Refer to the attachment for diagram

In Δ ADE

$\implies \tt \tan(60) = \frac{60}{DE}$

$\implies \tt \sqrt{3} = \frac{60}{DE}$

$\implies \tt DE = \frac{60}{ \sqrt{3} }$

$\implies \tt DE = 20 \sqrt{3} \: \: m$

Now , in Δ BCE

$\implies \tt \tan(30) = \frac{60}{CE}$

$\implies \tt \frac{1}{ \sqrt{3} } = \frac{60}{CD + 20 \sqrt{ 3} }$

$\implies \tt CD + 20 \sqrt{ 3 } = 60 \sqrt{3}$

$\implies \tt CD = 20 \sqrt{3} (3 - 1)$

$\implies \tt CD = 40 \sqrt{3}$

It is given that , the time taken by a bird to travel point D to C is 2 sec

Thus ,

$\implies \tt Speed = \frac{40 \sqrt{3} }{2}$

$\implies \tt Speed = 20 \sqrt{ 3} \: \: m {s}^{ - 1}$

★ The speed of flying bird is 20√3 m/s