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The shadow of a tower standing on a level ground is found to be $40 m$ longer when the Sun's altitude is $30 °$ than when it was $60 ° .$ then the height of the toweris, [Given $3 = 1.732$ ]

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24.64 m

34.64 m

44.64 m

54.64 m

answer is B.

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Detailed Solution

Given that the shadow of a tower standing on a level ground is found to be

40 m

longer when the Sun's altitude is

30 °

than when it was

60 ° .

We draw the diagram depicting the situation.
Question Image

We know,

tanθ = \frac{Perpendicular}{Base}

\tan 30 ° = \frac{1}{\sqrt{3}}

\tan 60 ° = \sqrt{3}

Now in triangle

ΔABD

\tan 60 ° = \frac{AB}{BD}

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

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

Now, in

ΔABC

\tan 30 ° = \frac{AB}{BC}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 40}

\Rightarrow x + 40 = \sqrt{3} h

\Rightarrow \frac{h}{\sqrt{3}} + 40 = \sqrt{3} h . . . . . . . . . . . from (1)

\Rightarrow h + 40 \sqrt{3} = 3 h

\Rightarrow 2 h = 40 \sqrt{3}

\Rightarrow h = 20 \sqrt{3}

\Rightarrow h = 34.64 m

The height of the tower is 34.64 m
Therefore, option 2 is correct.

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