Question

The angle of elevation of the top of the tower at a point on the ground 50 m away from the foot of the tower is $45^o$. Then the height of the tower is-
$
  {\text{A}}.\;50\sqrt 3 \;{\text{m}} \\
  {\text{B}}.\;50\;{\text{m}} \\
  {\text{C}}.\;\dfrac{{50}}{{\sqrt 2 }}\;{\text{m}} \\
  {\text{D}}.\;\dfrac{{50}}{{\sqrt 3 }}\;{\text{m}} \\
$

Answer 1

Hint: The angle of elevation is the angle above the eye level of the observer towards a given point. The angle of depression is the angle below the eye level of the observer towards a given point. The tangent function is the ratio of the opposite side and the adjacent side.

Complete step-by-step answer:

Let the height of the tower CD be h. Let E be the point 50 m away from the base of the tower CD. The angle of elevation at E is given as $45^o$. We will apply trigonometric formulas in triangle DCE.

In$ \vartriangle $DCE, 

 $ \tan {45^{\text{o}}} = \dfrac{{DC}}{{CE}} $

 $ 1 = \dfrac{{\text{h}}}{{50}} $

$  {\text{h}} = 50\;{\text{m}} $

This is the height of the tower, the correct option is B.

Note: In such types of questions, it is important to read the language of the question carefully and draw the diagram step by step correctly. When the diagram is drawn, we just have to apply basic trigonometry to find the required answer.