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A bird is sitting on the top of a tree, which is $80 m$ high. The angle of the elevation of the bird, from a point on the ground is $45 °$ . The bird flies away from the point of observation horizontally and remains at a constant height. After $2$ seconds, the angle of elevation of the bird from the point of observation becomes $30 °$ . Find the speed of the bird

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45 (\sqrt{3} - 1)

40 (\sqrt{4} - 1)

40 (\sqrt{3} - 1)

30 (\sqrt{3} - 1)

answer is C.

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Detailed Solution

It is given that a bird is sitting on the top of a tree, which is

80 m

high and angle of the elevation of the bird from a point on the ground is

45 ° .

The bird flies away but remains at a constant height. After

2

seconds, the angle of elevation of the bird from the point of observation becomes

30 °

.
According to the given data the figure has been drawn below.
Question Image

Let, AC be the tree and AC=DE=80m and CE=x.
In

Δ A B C,

tan θ = P e r p e n d i c u l a r b a s e

\tan 45 ° = \frac{AC}{BC}

1 = \frac{80}{BC}

∵ tan 45 ° = 1

BC = 80 m

Now,
In

Δ B D E,

tan θ = P e r p e n d i c u l a r b a s e

\tan 30 ° = \frac{DE}{BE}

\Rightarrow \tan 30 ° = \frac{DE}{BC + CE}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{80}{80 + x}

∵ tan 30 ° = 13

\Rightarrow 80 + x = 80 \sqrt{3}

\Rightarrow x = 80 \sqrt{3} - 80

Now,

x = CE = 80 (\sqrt{3} - 1) cm

.
The bird travelled

80 (\sqrt{3} - 1) m

.
Speed of bird,
Speed=

\frac{Distance}{time}

Substituting distance =

80 (\sqrt{3} - 1) m

and time= 2 sec in above formula we get

\Rightarrow \frac{80 (\sqrt{3} - 1)}{2}

\Rightarrow 40 (\sqrt{3} - 1) m / s

So, the speed of the bird is

40 (\sqrt{3} - 1) m / s

.
So, the correct option is 3.

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