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The shadow of tower standing on a level ground is $x m$ long ,when the Sun's altitude is $30 °$ , while it is $y m$ long when the altitude is $60 °$ . If the height of the tower is $45 \frac{\sqrt{3}}{2} m$ , then $x - y$ is

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45 m

45 \sqrt{3} m

\frac{45}{\sqrt{3}} m

45 \frac{\sqrt{3}}{2} m

answer is A.

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Detailed Solution

It is given that the shadow of tower standing on a level ground is

x m

long when the Sun's altitude is

30 °

.
Now it becomes

y m

long when the altitude is

60 °

.
According to the given data the figure has been drawn below.
Question Image

△ CAD,

tan θ = P e r p e n d i c u l a r b a s e

\tan 30 ° = \frac{45 \sqrt{3} / 2}{x}

\Rightarrow x = \frac{45 \sqrt{3}}{2 \times \frac{1}{\sqrt{3}}}

(\tan 30 ° = \frac{1}{\sqrt{3}})

\Rightarrow x = \frac{135}{2} m

……..(1)
And in

△ BAD

tan θ = P e r p e n d i c u l a r b a s e

\tan 60^{\circ} = \frac{\frac{45 \sqrt{3}}{2}}{y}

\Rightarrow \sqrt{3} = \frac{\frac{45}{2} \sqrt{3}}{y}

(\tan 60 ° = \sqrt{3})

\Rightarrow 1 = \frac{45}{2 y}

\Rightarrow y = \frac{45}{2}

…….(2)
From eq. (1) and (2),

x - y

will be

\Rightarrow \frac{135}{2} - \frac{45}{2}

\Rightarrow \frac{135 - 45}{2}

\Rightarrow \frac{90}{2}

\Rightarrow 45

So the value of

x - y

is 45 m.
So the correct option is 1.

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