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The shadow of a vertical tower on level ground increases by $20$ m, when the altitude of the sun changes from angle of elevation $60^{o}$ to $45^{o}$ . Find the height of the tower.

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47.32

47

48

49.56

answer is A.

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Detailed Solution

It is given that the shadow of a vertical tower on level ground increases by

20

m, when the altitude of the sun changes from angle of elevation

60^{o}

45^{o}

Let the height of the tower be

h

.
In right

ΔOBC,

\tan 60^{0} = \frac{h}{x}

⇒

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

⇒

h = \sqrt{3} x

....................(1)
In right

ΔOAC

\tan 45^{0} = \frac{h}{AB + BC}

⇒

\tan 45^{0} = \frac{h}{20 + x}

⇒

h = 20 + x

..............(2)
From (1) & (2), we get,
⇒

\sqrt{3} x = 20 + x

⇒

x = \frac{20}{\sqrt{3 - 1}}

⇒

x = \frac{20 \sqrt{3 + 1}}{2}

⇒

x = 10 (\sqrt{3} + 1)

⇒

x = 27.32

m
⇒

h = 20 + x

⇒

h = 20 + 27.32

⇒

h = 47.32 m

Thus, the height of the tower is

47.32 m

.
Hence, the correct option is 1.

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