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The shadow of the tower becomes 60 m longer when the altitude of the sun changes from $45^{\circ} to 30^{\circ}$ . Then the height of the tower is:

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20 (\sqrt{3} + 1) m

24 (\sqrt{3} + 1) m

30 (\sqrt{3} + 1) m

30 (\sqrt{3} - 1) m

answer is C.

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Detailed Solution

It is given that the shadow of the tower becomes 60 m longer when the altitude of the sun changes from

45^{\circ} to 30^{\circ}

. We have to find the height of the tower.
We use

tanθ = \frac{perpendicular}{base} .

ΔTOW

\tan 45^{o} = \frac{TW}{OW}

1 = \frac{TW}{OW}

[since

\tan 45^{o}

=1]

\Rightarrow x = h

ΔTXW

\tan 30^{o} = \frac{TW}{XW}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{60 + h}

[since

\tan 30^{o}

\frac{1}{\sqrt{3}}

]

\Rightarrow 60 + h = \sqrt{3} h

\Rightarrow 60 = h (\sqrt{3} - 1)

\Rightarrow h = \frac{60}{\sqrt{3} - 1}

\Rightarrow h = \frac{60 (\sqrt{3} + 1)}{3 - 1}

\Rightarrow h = 30 (\sqrt{3} + 1) m

Hence, the height of the tower is

30 (\sqrt{3} + 1) m

.
Therefore, the correct option is 3.

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