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From the top of a cliff $25 m$ high, the angle of elevation of the top of tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower.

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25 m

50 m

\frac{25}{\sqrt{2}} m

25 \sqrt{2} m

answer is B.

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Detailed Solution

Given that a cliff is 25 m high and the angle of elevation of the top of tower is equal to the angle of depression of the foot of the tower.
Question Image

Take into account DCB with a straight angle at C.

\tan x^{\circ} = \frac{BC}{CD}

\Rightarrow \tan x^{\circ} = \frac{25}{CD} . . . (1)

Now, consider

ΔDCE

right angled at C,

\tan x^{\circ} = \frac{CE}{CD} . . . (2)

Equating equation (1) & (2),

\Rightarrow \frac{25}{CD} = \frac{CE}{CD}

\Rightarrow CE = 25

∵

BE is the height of a tower such that BE=BC+CE.
As, BC = AD = 25

\Rightarrow BE = 25 + CE

∴

Total height of tower is (25+25) m

\Rightarrow

Total height of tower

= 50 m

So, the height of the tower is 50 m.
Hence, the correct option is 2.

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