Question

(ii) A peacock is sitting on the tree and observes its prey on

the ground. It makes an angle of depression of 22° to
catch the prey. The speed of the peacock was observed to
be 10 km/hr and it catches its prey in 1 min 12 seconds.
At what height was the peacock on the tree?
(cos 22° = 0.927, sin 22° = 0.374, tan 22° = 0.404)
​

Answer 1

74.8 m

Explanation:

We are given that

`\theta=22^(\circ)`

Speed of peacock,v=10 km/h

Time,t=1 min 12 sec=60+12=72 s=
`(72)/(3600)=0.02 h`

1 hour=3600 s

Distance=vt=
`10* 0.02=0.2 km=0.2* 1000=200 m`

Using 1 km=1000 m

We know that

`sin\theta=(Perpendicular\;side)/(hypotenuse)`

Using the formula

`sin22=(h)/(200)`

`h=200sin22=200* 0.374=74.8 m`

Hence, the peacock was at height of 74.8 m on the tree.