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The shadow of the tower standing on a level ground is found to be $60$ metres longer when the sun's altitude is $30^{o}$ than when it is $45^{o}$ . Find the height of the tower.

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60 m

30 m

60 \sqrt{3} m

30 (\sqrt{3} + 1) m

answer is D.

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Detailed Solution

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

60

metres longer when the sun's altitude is

30^{o}

than when it is

45^{o}

.
Let XY be the height of tower, YZ (x) be the shadow when sun's altitude is

30^{o}

and AY be the length of shadow when sun's altitude is

45^{o}

△ XYZ

\tan 45^{\circ} = \frac{XY}{YZ}

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

\Rightarrow XY = x

cm
In

△ XYA

\tan 30^{\circ} = \frac{XY}{AY}

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

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

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

Now, rationalise the fraction to get the value of x.

x = \frac{60}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}

\Rightarrow x = \frac{60 (\sqrt{3} + 1)}{(\sqrt{3})^{2} - (1)^{2}}

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

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

m
Height of tower is

30 (\sqrt{3} + 1)

m.
Hence, the correct option is 4.

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