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An observer finds the angle of elevation of the top of a tower from certain point is $30 ° .$ If the observer moves $20 m$ towards the tower, the angle of elevation of the top increases by $15 ° .$ Then the height of the tower is:

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10 (3 + 1)

10 (3 + 3)

10 (2 + 1)

13 (3 + 1)

answer is A.

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Detailed Solution

Given that, an observer finds the angle of elevation of the top of a tower from certain point is

30 ° .

If the observer moves

20 m

towards the tower, the angle of elevation of the top increases by

15 ° .

Let PR=h be the height of the tower now,
the observer is standing at point Q, such that the distance between the observer and the tower is

Q R = 20 + x m

,
where

Q R = Q S + S R = 20 + x

∠ PQR = 30 °

∠ PSR = θ

We know,

tanθ = \frac{Perpendicular}{Base}

\tan 60 ° = \sqrt{3}

\tan 30 ° = \frac{1}{\sqrt{3}}

Now in triangle PQR,

\tan 30 ° = \frac{h}{20 + x}

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

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

\Rightarrow x = \sqrt{3} h - 20 . . . . . . . . . . . . . . (1)

Now, in triangle PSR,

tan θ = h x

since it is been given that the angle of elevation increases by

15 °

if the observer moves 20m,
Now we have,

θ = 30 ° + 15 °

\Rightarrow θ = 45 °

So,

\tan 45^{\circ} = \frac{h}{x}

\Rightarrow 1 = \frac{h}{x}

\Rightarrow h = x

Now substituting the value of

x

in equation 1, we get,

⇒ h = 3 h − 20 ⇒ 3 h − h = 20 ⇒ h (3 − 1) = 20 ⇒ h = 20 3 − 1

After rationalizing the denominator, we have,

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

\Rightarrow h = \frac{20 (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)}

\Rightarrow h = \frac{20 (\sqrt{3} + 1)}{3 - 1}

\Rightarrow h = \frac{20 (\sqrt{3} + 1)}{2}

\Rightarrow h = 10 (\sqrt{3} + 1)

m
Therefore, the required height of the tower is

10 (\sqrt{3} + 1

) m.
Therefore, option 1 is correct.

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