Name: Heights and Distances: Illustration 4
Uploaded: 2019-07-19
Duration: 25 min 42 s
Description: In this video, we will find the height of the tower and building using the application of trigonometry using the terminologies related to the chapter. Let's ...

Question

A bird is sitting on the top of a $80 m$ high tree. From a point on the ground, the angle of elevation of the bird is $45 °$ . The bird flew away horizontally in such a way that it remained at a constant height from the ground. After $2$ seconds, the angle of elevation of the bird from the same point is $30 °$ . Find the speed of flying of the bird. (Take $\sqrt{3} = 1.732$ )

Accepted Answer

From the figure, we have

$BC = ED = 80 m$

$CD = BE = d$

In $△ ABC, \tan 45 ° = \frac{BC}{AB}$

$\Rightarrow 1 = \frac{BC}{AB}$

$\Rightarrow AB = BC = 80 m$

Now, in $△ AED, \tan 30 ° = \frac{ED}{AE} = \frac{BC}{AB + BE}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{80}{80 + CD} = \frac{80}{80 + d}$

$\Rightarrow d = 80 (\sqrt{3} - 1) m$

$Speed = \frac{Distance}{Time}$

Since, the bird takes $2$ seconds to cover $80 (\sqrt{3} - 1) m$ distance, its speed $= \frac{80 (\sqrt{3} - 1)}{2} = 40 (\sqrt{3} - 1) m / s$

$40 (\sqrt{3} - 1) m / s = (40 \times 0.732) = 29.28 m / s$ $[∵ \sqrt{3} = 1.732]$

Therefore, the speed of the flying of the bird $= 29.28 m / s$

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