Question

A tower subtends an angle of $30^o$ at a point on the same level at its foot. At a second point h metres above the first, the depression of the foot of the tower is $60^o$. The height of the tower is-
$
  {\text{A}}.\;\dfrac{{\text{h}}}{2}{\text{m}} \\
  {\text{B}}.\;\sqrt 3 {\text{h}}\;{\text{m}} \\
  {\text{C}}.\;\dfrac{{\text{h}}}{3}\;{\text{m}} \\
  {\text{D}}.\;\dfrac{{\text{h}}}{{\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:

We have to find the height of the tower CD. Let E be the point on the ground where the tower subtends $30^o$, and F be the point at a height of h m above E. Let CE = x. We will apply trigonometric formulas in triangles DCE and FEC.

$

  In\;\vartriangle DCE, \\

  \tan {30^{\text{o}}} = \dfrac{{DC}}{{CE}} \\

  \dfrac{1}{{\sqrt 3 }} = \dfrac{{DC}}{{\text{x}}} \\

  DC = \dfrac{{\text{x}}}{{\sqrt 3 }} \\

$
$
  \angle CFG\;and\angle FCE\;are\;alternate\;interior\;angles\;and\;are\;equal. \\

  In\;\vartriangle FCE, \\

  \tan {60^{\text{o}}} = \dfrac{{EF}}{{CE}} \\

  \sqrt 3 = \dfrac{{\text{h}}}{{\text{x}}} \\

  {\text{x}} = \dfrac{{\text{h}}}{{\sqrt 3 }} \\

$
Substituting this value of x in the other equation,

$

  DC = \dfrac{{\text{h}}}{{\sqrt 3 \left( {\sqrt 3 } \right)}} \\

  DC = \dfrac{{\text{h}}}{3} \\

$



Hence, the correct option is C.


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.