From the top of a cliff 25m high the angle of elevation of a tower is found to b

Question

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

Options:

A . 25m

B . 75m

C . 50m

D . 100m

Answer: Option C
:
C

From The Top Of A Cliff 25m High The Angle Of Elevation Of A...

In triangle ABE,

t a n θ_{2}

\frac{A B}{B E}

\frac{25}{B E}

In triangle ADC,

t a n θ_{1}

\frac{C D}{A D}

We know,

θ_{1}

θ_{2}

\Rightarrow

t a n θ_{1}

t a n θ_{2}

\frac{C D}{A D}

\frac{25}{B E}

\Rightarrow

CD = 25 [Since AD = BE]
DE =AB = 25m

\Rightarrow

Height of tower = CD+ DE
= 25+25 = 50m

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