Question

A peacock is sitting on a 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 10km/he and it catches its prey in 1min 12 secs. At what height was the peacock on the tree?
( cos 22° = 0.927 , sin 22° = 0.374 , tan 22° = 0.404

Answer 1

Thank you for asking this question. Here is your answer:

∠DCA = ∠BAC = 30°

Speed of peacock is 300 m/min

= 300 m/60s

= 5 m/s

The prey is caught in 12 seconds.

Distance traveled by peacock to catch the prey = x m/12 s

x = 60 m

AC = 60 m

Height of tree = BC

In ΔABC, Sin 30° = BC/AC = BC/60 = 1/2

BC = 60/2 = 30 m

The height of tree = 30m

If there is any confusion please leave a comment below.

Answer 2

∠DCA = ∠BAC = 30°

Speed of peacock is 300 m/min

= 300 m/60s

= 5 m/s

The prey is caught in 12 seconds.

Distance traveled by peacock to catch the prey = x m/12 s

x = 60 m

AC = 60 m

Height of tree = BC

In ΔABC, Sin 30° = BC/AC = BC/60 = 1/2

BC = 60/2 = 30 m

The height of tree = 30m

If there is any confusion please leave a comment below.