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

$\bold{\large{\underline{\underline{\rm{\red{AnsWer:}}}}}}$

$\bold{\boxed{\tt{\blue{Peacock\:was\:at\:a\:height\:of\:74.8\:m}}}}$

Given :

• Angle of depression = 22°
• Speed of the peacock = 10 km/hr
• Time taken by the peacock to catch the prey = 1 min 12 seconds = 60 sec + 12 sec = 72 seconds.
• Cos 22° = 0.927
• Sin 22° = 0.374
• Tan 22° = 0.404

To Find :

• Height at which the peacock was initially on the tree.

Solution :

AC is the distance covered by the peacock to reach out to the prey.

We can calculate AC using the relation between speed, time and distance.

$\bold{\sf{\large{\red{\boxed{Distance,AC\:=\:Speed\:\times\:Time}}}}}$

Block in the available data,

➟ $\mathtt{AC\:=\:10\:\times\:{\dfrac{72}{3600}}}$

➟ $\mathtt{AC\:=\:{\dfrac{72}{360}}}$

➟ $\mathtt{AC=\:{\dfrac{1}{5}}}$

➟ $\mathtt{AC\:=\:0.2}$

AC = 0.2 km

Change km into m :

1 km = 1000 m

➟ $\mathtt{0.2\:\times\:1000}$

➟ $\mathtt{200\:m}$

$\sf{\therefore{\underline{AC\:=\:200\:m}}}$

CD is the height at which the peacock was initially.

Let CD = x m.

In right angled Δ CDA,

$\theta$ = 22°

➟ $\mathtt{Sin\:\theta\:=\:{\dfrac{Opposite\:side\:to\:\theta}{Hypotenuse}}}$

➟ $\mathtt{Sin\:22\:\degree\:=\:{\dfrac{CD}{200}}}$

➟ $\mathtt{0.374\:=\:{\dfrac{x}{200}}}$

➟ $\mathtt{0.374\:\times\:200\:=\:x}$

➟ $\mathtt{74.8\:m\:=\:x}$

$\sf{\therefore{\underline{Height\:at\:which\:peacock\:was\:on\:the\:tree\:=\:CD\:=\:74.8\:m}}}$

Answer 2

Answer:

$\bold{\boxed{\tt{\orange{Peacock\:was\:at\:a\:height\:of\:74.8\:m}}}}$

• Given:-

•Angle of depression = 22°

•Speed of the peacock = 10 km/hr

•Time taken by the peacock to catch the

prey = 1 min 12 seconds = 60 sec + 12 sec = 72 seconds.

•Cos 22° = 0.927

•Sin 22° = 0.374

•Tan 22° = 0.404

• To Find:-

Height at which the peacock was initially on the tree.

Solution :

AC is the distance covered by the peacock to reach out to the prey.

We can calculate AC using the relation between speed, time and distance.

$\bold{\sf{\large{\red{\boxed{Distance,AC\:=\:Speed\:\times\:Time}}}}}$

• Block in the available data,

➟ $\mathtt{AC\:=\:10\:\times\:{\dfrac{72}{3600}}}$

➟ $\mathtt{AC\:=\:{\dfrac{72}{360}}}$

➟ $\mathtt{AC=\:{\dfrac{1}{5}}}$

➟ $\mathtt{AC\:=\:0.2}$

• Change km into m :

1 km = 1000 m

➟ $\mathtt{0.2\:\times\:1000}$

➟ $\mathtt{200\:m}$

$\sf{\therefore{\underline{AC\:=\:200\:m}}}$

CD is the height at which the peacock was initially.

Let CD = x m.

In right angled Δ CDA,

$\thetaθ = 22°$

➟ $\mathtt{Sin\:\theta\:=\:{\dfrac{Opposite\:side\:to\:\theta}{Hypotenuse}}}$

➟ $\mathtt{Sin\:22\:\degree\:=\:{\dfrac{CD}{200}}}$

➟ $\mathtt{0.374\:=\:{\dfrac{x}{200}}}$

➟ $\mathtt{0.374\:\times\:200\:=\:x}$

➟ $\mathtt{74.8\:m\:=\:x}$

$\sf{\therefore{\underline{Height\:at\:which\:peacock\:was\:on\:the\:tree\:=\:CD\:=\:74.8\:m}}}$∴