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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 $ 22{}^\circ $ to catch the prey.

Now, letâ€™s draw a diagram by using the given information.

We have to find the height of the tree $ AB $.

Let us assume that peacock is sitting at the point $ A $ and prey is on the ground at point $ C $ . Now, as given in the question It makes an angle of depression of $ 22{}^\circ $ to catch the prey so $ \angle CAD=22{}^\circ $.

Also, $ \angle ACB=22{}^\circ $ as $ \angle CAD $ and $ \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/hr $ and it catches its prey in $ \text{1 min 12 seconds} $. A peacock travels a path $ AC $ to reach the prey, so the distance travelled by the peacock to reach the prey will be calculated by using the formula $ \text{Speed=}\dfrac{\text{Distance}}{\text{Time}} $

We have given the speed of the peacock is $ 10km/hr $ and time taken to reach the prey is $ \text{1 min 12 seconds} $ .

Then, distance will be $ \text{Distance=speed}\times \text{time} $

$ \text{Distance=}10km/hr\times \text{1 min 12 seconds} $

Now, we have to convert the units. We know that

$ \begin{align}

& 1\text{ minute = 60 seconds} \\

& \text{1 km/hr = }\dfrac{5}{18}m/\sec \\

\end{align} $

So, we get

$ \begin{align}

& \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{align} $

In the figure drawn the length of $ AC=200 $

Now, let us consider right angle triangle $ \Delta ABC, $

We know that $ \sin \theta =\dfrac{\text{Perpendicular}}{\text{hypotenuse}} $

We have $ \theta =22{}^\circ $ , substituting the values we get

$ \begin{align}

& \sin \theta =\dfrac{\text{AB}}{\text{AC}} \\

& \sin 22{}^\circ =\dfrac{AB}{200} \\

\end{align} $

Now, we have given in the question $ \sin 22{}^\circ =0.374 $

$ \begin{align}

& 0.374=\dfrac{AB}{200} \\

& 0.374\times 200=AB \\

& AB=74.8 \\

\end{align} $

So, the height of the tree is $ 74.8\text{ m} $ .

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