From the top of a cliff 25 m high the angle of elevation of a tower
Question:

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

(a) 25 m

(b) 50 m

(c) 75 m

(d) 100 m

Solution:

Given that: height of cliff is m and angle of elevation of the tower is equal to angle of depression of foot of the tower that is.

Now, the given situation can be represented as,

Here, D is the top of cliff and BE is the tower.

Let $C E=h, A B=x$. Then, $A B=D C=x$

Here, we have to find the height of the tower BE.

So, we use trigonometric ratios.

In a triangle ABD,

$\Rightarrow \tan \theta=\frac{A D}{A B}$

$\Rightarrow \tan \theta=\frac{25}{x}$…………..(1)

Again in a triangle,

$\tan \theta=\frac{C E}{C D}$

$\Rightarrow \tan \theta=\frac{h}{x}$

$\Rightarrow \frac{25}{x}=\frac{h}{x}$ $[$ Using (1) $]$

$\Rightarrow h=25$

Thus, height of the tower = BE = BC + CE = (25 + 25) m = 50 m

Hence, the correct option is .