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From a point on the ground, the angle of elevation of the top of a tower is observed to be $60 °$ From a point $40 m$ vertically above the first point of observation, the angle of elevation of the top of the tower is $30 °$ ,then the height of the tower and its horizontal distance from the point of observation is

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203 m

, 50m

203 m

, 60 m

203 m

, 40m

203 m

, 30m

answer is D.

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Detailed Solution

Given that from a point on the ground, the angle of elevation of the top of a tower is observed to be

60 °

From a point

40 m

vertically above the first point of observation, the angle of elevation of the top of the tower is

30 °

.
Consider the diagram,
Question Image

We know,

tanθ = \frac{Perpendicular}{Base}

\tan 60 ° = \sqrt{3}

\tan 30 ° = \frac{1}{\sqrt{3}}

,
In triangle QAP,

\tan 60 ° = \frac{QP}{AP}

\Rightarrow \tan 60 ° = \frac{x + 40}{AP}

In triangle QBP,

\tan 30 ° = \frac{QR}{BR}

\Rightarrow \tan 30 ° = \frac{x}{BR}

Therefore,

AP = x + 40 tan 60 ° BR = x tan 30 °

As AP and BR are equal so equate the expression of both to each other and find x.

x + 40 tan 60 ° = x tan 30 ° ⇒ x + 40 3 = x 3 ⇒ 2 x = 40 ⇒ x = 20 m

Now,

h = 20 + 40 ⇒ h = 60 m

Again,

d = \frac{20}{\tan 30 °}

\Rightarrow d = \frac{20}{\frac{1}{\sqrt{3}}}

\Rightarrow d = 20 \sqrt{3} m

Distance between point and tower

203 m

and the height of the tower is 60 m.
Therefore, option 4 is correct.

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