Question

From the top of a cliff 20 m high the angle of elevation of the top of the Tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is
A) 20
B) 40
C) 60
D) 80

Answer 1

Hint:
We will find the height of the tower by applying concepts of trigonometry. We will draw a figure representing the situation in the question. We will find the height of the tower by finding the ratio \[\tan \theta \] in the 3 triangles of the figure that we have drawn.
Formulas used: We will use the formula for the measure of the tangent of an angle \[\theta \] in a right-angled triangle is the ratio of the triangle’s perpendicular and base, \[\tan \theta = \dfrac{P}{B}\].

Complete step by step solution:
 

The cliff is represented by the line segment \[AB\] and the tower is represented by the line segment \[CD\] . We will assume that the angle of elevation and the angle of depression is \[\theta \] .
So,
\[\angle CAE = \angle EAD = \theta \]
We will assume that the horizontal distance of the tower from the cliff is \[x\] . So,
\[BD = AE = x\]
We can see from the figure that the length of \[ED\] is also 20 m.
Line segments \[EA\] and \[DB\] are parallel to each other and \[AD\] is acting as a transversal. So, the measure of \[\angle EAD\] will be equal to the measure of angle \[\angle ADB\] because they are alternate interior angles. We know that the alternate interior angles are equal, therefor,
\[\angle EAD = \angle ADB = \theta \]
Using the formula for \[\tan \theta \], in triangle \[ABD\], we get
\[\tan \theta = \dfrac{{AB}}{{DB}}\]
Substituting \[AB = 20\] and \[DB = x\] in the above equation, we get
\[ \Rightarrow \tan \theta = \dfrac{{20}}{x}\]……………\[\left( 1 \right)\]
Using the formula for \[\tan \theta \], in triangle \[AED\], we get
\[\tan \theta = \dfrac{{DE}}{{AE}}\]
Substituting \[DE = 20\] in the above equation, we get
\[ \Rightarrow \tan \theta = \dfrac{{20}}{{AE}}\]………………..\[\left( 2 \right)\]
We will now compare equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\]. So, we get
\[\dfrac{{20}}{{AE}} = \dfrac{{20}}{x}\]
On cross multiplication, we get
\[ \Rightarrow 20AE = 20x\]
\[ \Rightarrow {\rm{ }}AE = x\]………………..\[\left( 3 \right)\]
Using the formula for \[\tan \theta \], in triangle \[AEC\], we get
\[\tan \theta = \dfrac{{CE}}{{EA}}\]
Substituting the value of \[EA\] from equation \[\left( 3 \right)\], we get
\[ \Rightarrow \tan \theta = \dfrac{{CE}}{x}\]…………………\[\left( 4 \right)\]
Comparing equation \[\left( 1 \right)\] and equation \[\left( 4 \right)\], we get
\[\begin{array}{l}\dfrac{{CE}}{x} = \dfrac{{20}}{x}\\ \Rightarrow {\rm{ }}CE = 20\end{array}\]
We know that the length of \[CE\] and the length of \[ED\] is 20 m each.
The length of \[CD\] will be:
\[CD = CE + ED\]
Substituting the value of \[CE\] and \[ED\], we get
\[\begin{array}{l} \Rightarrow CD = 20 + 20\\ \Rightarrow CD = 40\end{array}\]
The length of \[CD\] is 40 m. This means that the height of the tower is 40 m.

$\therefore $ Option B is the correct option.

Note:
We can also find the length of \[CE\] using the concept of congruence in the triangle \[AED\] and triangle \[AEC\]. We will use Angle-Side-Angle congruence and criterion and prove that the length of \[CE\] and \[AD\] is the same. We know that the length of \[DE\] is 20 m (through symmetry). We will add the lengths of \[CE\] and \[AD\] to find the height of the tower.