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A vertical tower of height $50$ m high stands on a sloping ground. The bottom of the tower is at the same level as the middle point of a vertical flagpole. From the top of the tower, the angles of depression of the top and bottom of the flagpole are $15^{0}$ and $45^{0}$ respectively. What is the height of the flagpole?

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\frac{50}{\sqrt{3}} m

50 \sqrt{3} m

\frac{100}{\sqrt{3}} m

100 \sqrt{3} m

answer is C.

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Detailed Solution

It is given that a vertical tower of height

50

m high stands on a sloping ground. The bottom of the tower is at the same level as the middle point of a vertical flagpole. From the top of the tower, the angles of depression of the top and bottom of the flagpole are

15^{0}

and

45^{0}

respectively.
Question Image

△ ABC,

\tan 15 ° = \frac{50 - x}{a}

a = \frac{50 - x}{\tan 15 °}

… (1)
In

△ ADE,

\tan 45 ° = \frac{50 + x}{a}

⇒

a = 50 + x

⇒

\frac{(50 - x)}{\tan 15 °} = 50 + x

… from (1).
⇒

50 - x = \tan 15 ° (50 + x)

⇒

50 - x = \frac{(\sqrt{3} - 1)}{(\sqrt{3} + 1)} (50 + x) . . . (\tan 15 ° = \frac{(\sqrt{3} - 1)}{(\sqrt{3} + 1)})

.
⇒

(50 - x) (\sqrt{3} + 1) = (\sqrt{3} - 1) (50 + x)

⇒

50 \sqrt{3} - \sqrt{3} x + 50 - x = 50 \sqrt{3} - 50 + \sqrt{3} x - x

⇒

2 \sqrt{3} x = 100

⇒

x = \frac{50}{\sqrt{3}}

⇒

2 x = \frac{100}{\sqrt{3}}

∴

Height of flagpole is

\frac{100}{\sqrt{3}} m

.
Hence, the correct option is 3.

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