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The angle of elevation of a tower from a point on the ground is $30 °$ . At a point on the horizontal line passing through the foot of the tower and $100$ meters nearer to it, the angle of elevation is found to be $60 °$ , then the height of the tower is

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50 \sqrt{3}

\frac{50}{\sqrt{3}}

\frac{100}{\sqrt{3}}

100 \sqrt{3}

answer is A.

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Detailed Solution

It is given that the angle of elevation of the tower from a point on the ground is

30 °

. At a point which is

100

meters nearer to it, the angle of elevation is found to be

60 °

from that point.
Let

= x m

BC = 100 m

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

From

△ ACD,

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

\tan 60 ° = \frac{AD}{CD}

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

∵ tan 60 ° = 3

\Rightarrow AD = x \sqrt{3} . . . . . . . . . . (i)

From

△ ABD,

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

\tan 30 ° = \frac{AD}{BD}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{AD}{x + 100}

∵ tan 30 ° = 13

\Rightarrow \sqrt{3} AD = x + 100

\Rightarrow 3 x = x + 100

[By (i)]

\Rightarrow x = 50

From the equation (i) we get,
AD

= 50 \times \sqrt{3} m

= 50 \sqrt{3} m

Hence the height of the tower is

50 \sqrt{3} m .

So the correct option is 1.

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