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Question: A peacock is sitting on the tree and observes its prey on the ground. It makes an angle of depressio...

A peacock is sitting on the tree and observes its prey on the ground. It makes an angle of depression of 2222{}^\circ to catch the prey. The shadow of the peacock was observed to be 10km/hr10km/hr and it catches its prey in 1 min 12 seconds\text{1 min 12 seconds} . At what height was the peacock on the tree? (cos22=0.927,sin22=0.374,tan22=0.404)\left( \cos 22{}^\circ =0.927,\sin 22{}^\circ =0.374,\tan 22{}^\circ =0.404 \right)

Explanation

Solution

Hint : First we draw a diagram, then we find the distance travelled by the peacock to catch the prey as given that the shadow of the peacock was observed to be 10km/hr10km/hr and it catches its prey in 1 min 12 seconds\text{1 min 12 seconds} . Then, use this distance as the length of a side of the triangle and use trigonometric property to find the height of the tree.

Complete step-by-step answer :
We have given that a peacock is sitting on the tree and observes its prey on the ground. It makes an angle of depression of 2222{}^\circ to catch the prey.
Now, let’s draw a diagram by using the given information.

We have to find the height of the tree ABAB.
Let us assume that peacock is sitting at the point AA and prey is on the ground at point CC . Now, as given in the question It makes an angle of depression of 2222{}^\circ to catch the prey so CAD=22\angle CAD=22{}^\circ.
Also, ACB=22\angle ACB=22{}^\circ as CAD\angle CAD and ACB\angle ACB are alternate angles and angle of depression is equal to angle of elevation.
Also, it is given that the shadow of the peacock was observed to be 10km/hr10km/hr and it catches its prey in 1 min 12 seconds\text{1 min 12 seconds}. A peacock travels a path ACAC to reach the prey, so the distance travelled by the peacock to reach the prey will be calculated by using the formula Speed=DistanceTime\text{Speed=}\dfrac{\text{Distance}}{\text{Time}}
We have given the speed of the peacock is 10km/hr10km/hr and time taken to reach the prey is 1 min 12 seconds\text{1 min 12 seconds} .
Then, distance will be Distance=speed×time\text{Distance=speed}\times \text{time}
Distance=10km/hr×1 min 12 seconds\text{Distance=}10km/hr\times \text{1 min 12 seconds}
Now, we have to convert the units. We know that
1 minute = 60 seconds 1 km/hr = 518m/sec \begin{aligned} & 1\text{ minute = 60 seconds} \\\ & \text{1 km/hr = }\dfrac{5}{18}m/\sec \\\ \end{aligned}
So, we get
Distance=(10×518)×(60+12) Distance=5018×72 Distance=200 m \begin{aligned} & \text{Distance=}\left( 10\times \dfrac{5}{18} \right)\times \left( 60+\text{12} \right) \\\ & \text{Distance}=\dfrac{50}{18}\times 72 \\\ & \text{Distance}=200\text{ m} \\\ \end{aligned}
In the figure drawn the length of AC=200AC=200
Now, let us consider right angle triangle ΔABC,\Delta ABC,
We know that sinθ=Perpendicularhypotenuse\sin \theta =\dfrac{\text{Perpendicular}}{\text{hypotenuse}}
We have θ=22\theta =22{}^\circ , substituting the values we get
sinθ=ABAC sin22=AB200 \begin{aligned} & \sin \theta =\dfrac{\text{AB}}{\text{AC}} \\\ & \sin 22{}^\circ =\dfrac{AB}{200} \\\ \end{aligned}
Now, we have given in the question sin22=0.374\sin 22{}^\circ =0.374
0.374=AB200 0.374×200=AB AB=74.8 \begin{aligned} & 0.374=\dfrac{AB}{200} \\\ & 0.374\times 200=AB \\\ & AB=74.8 \\\ \end{aligned}
So, the height of the tree is 74.8 m74.8\text{ m} .

Note : The point to remember is that always keep the units the same for all variables. If units are not the same you have to convert the units. In this particular question one may be confused after reading the line that the shadow of the peacock was observed to be 10km/hr10km/hr and it catches its prey in 1 min 12 seconds\text{1 min 12 seconds} . But use this line to get the value of a side of the triangle, so you can use it to solve the trigonometric ratio and obtain an answer.