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

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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