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The horizontal distance between two towers is $30$ meters. From the foot of the first tower the angle of elevation of the top of the second tower is $60 °$ . From the top of the second tower the angle of depression of the top of the first is 30 $°$ . The height of the small tower is

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20 (\sqrt{3} + 1)

30 (\sqrt{3} - 1)

30 \sqrt{3}

30

answer is C.

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Detailed Solution

It is given that the horizontal distance between two towers is

30

meters and from the foot of the first tower the angle of elevation of the top of the second tower is

60 °

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

In the figure we get,
In

Δ C D E,

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

\tan 60 ° = \frac{h_{1}}{30}

…(1)
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 30 ° = \frac{h_{2} - h_{1}}{30}

…(2)

\Rightarrow \tan 30 ° = \frac{h_{2}}{30} - \frac{h_{1}}{30}

From (1) and (2) we get,

\Rightarrow \frac{h_{2}}{30} = \tan 60 ° + \tan 30 °

\Rightarrow \frac{h_{2}}{30} = \sqrt{3} + \frac{1}{\sqrt{3}}

∵ tan 30 ° = 13, tan 60 ° = 3

\Rightarrow \frac{h_{2}}{30} = \frac{3 + 1}{\sqrt{3}}

\Rightarrow \frac{h_{2}}{30} = \frac{4}{\sqrt{3}}

\Rightarrow h_{2} = \frac{30 \times 4}{\sqrt{3}}

\Rightarrow h_{2} = 40 \sqrt{3}

So the height of the big tower is

40 \sqrt{3}

.
From the equation (2) we get,

\tan 30 ° = \frac{h_{2}}{30} - \frac{h_{1}}{30}

\Rightarrow \tan 30 ° = \frac{40 \sqrt{3}}{30} - \frac{h_{1}}{30}

\Rightarrow \frac{40 \sqrt{3}}{30} - \tan 30 ° = \frac{h_{1}}{30}

\Rightarrow \frac{40 \sqrt{3}}{30} - \frac{1}{\sqrt{3}} = \frac{h_{1}}{30}

∵ tan 30 ° = 13

\Rightarrow \frac{4}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{h_{1}}{30}

\Rightarrow \frac{4 - 1}{\sqrt{3}} = \frac{h_{1}}{30}

\Rightarrow \frac{3}{\sqrt{3}} = \frac{h_{1}}{30}

\Rightarrow \sqrt{3} = \frac{h_{1}}{30}

\Rightarrow 30 \sqrt{3} = h_{1}

So the height of the small tower is

30 \sqrt{3}

m.
So the correct option is 3.

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