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The shadow of a tower standing on a level plane is found to be $40 m$ longer when Sun's elevation is $30 °$ than when it is $60 ° .$ ,then the height of the tower is

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203

103

20m

10m

answer is A.

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Detailed Solution

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

40 m

longer when Sun's elevation is

30 °

than when it is

60 ° .

Considering the figure,
Question Image

We know,

tanθ = \frac{Perpendicular}{Base}

\tan 30 ° = \frac{1}{\sqrt{3}}

In triangle ABC,

tan 60 0 = A B BC ∴ 3 = h B C ⇒ h= 3 B C

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

Now in triangle ABD,

\tan 30^{0} = \frac{AB}{BD}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{BD}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{BC + CD}

\Rightarrow BC + CD = h \sqrt{3}

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

\Rightarrow 40 = h (\sqrt{3} - \frac{1}{\sqrt{3}})

\Rightarrow 40 = h (\frac{3 - 1}{\sqrt{3}})

\Rightarrow 40 = h (\frac{2}{\sqrt{3}})

\Rightarrow h = 40 \times \frac{\sqrt{3}}{2}

\Rightarrow h = 20 \sqrt{3}

m
Therefore the height of the tower is

203

m.
Hence, option 1 is correct.

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