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The angle of elevation of the top of the tower at any point of the ground is $30 °$ and if moving $20 m$ towards the tower it becomes $60 °,$ then the height of the tower is

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10

10 \sqrt{3}

\frac{10}{\sqrt{3}}

30

answer is B.

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Detailed Solution

It is given the angle of elevation of the top of the tower at any point of the ground is

30 °

.
If one moves

20 m

towards the tower angle of elevation becomes 60

°

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

Angle of elevation of the top of a tower at

D = 30 ° .

Distance between

C

and

D (CD) = 20 m

and angle of elevation of the top of the tower at

C = 60 ° .

Let

AB

be the tower of height

h

and

x

be the distance between

A

and

C

i.e,

AC = x

.
In

△ ACB,

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

\tan 60 ° = \frac{BA}{AC}

= \frac{h}{x}

\Rightarrow \sqrt{3} = \frac{h}{x}

∵ tan 60 ° = 3

\Rightarrow h = \sqrt{3} x

...(1)
We also know that in

△ ABD,

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

\tan 30 ° = \frac{BA}{AD}

= \frac{BA}{AC + CD}

(from figure)

= \frac{h}{x + 20}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 20}

∵ tan 30 ° = 13

\Rightarrow \sqrt{3} h = x + 20

\Rightarrow \sqrt{3} h - 20 = x

...(2)
Substituting the value of x in the equation. (1), we get

h = \sqrt{3} (\sqrt{3} h - 20)

\Rightarrow h = 3 h - 20 \sqrt{3}

\Rightarrow - 2 h = - 20 \sqrt{3}

\Rightarrow h = 10 \sqrt{3}

m
So the height of the tower is

10 \sqrt{3}

m.
So the correct option is 2.

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