Question

The angles of depression of to ship from the top of the light house are $45^o$ and $30^o$ towards east. If the ships are 100 m apart, the height of the lighthouse is-
$
  {\text{A}}.\;\dfrac{{50}}{{\sqrt 3 + 1}}{\text{m}} \\
  {\text{B}}.\;\dfrac{{50}}{{\sqrt 3 - 1}}{\text{m}} \\
  {\text{C}}.\;50\left( {\sqrt 3 - 1} \right){\text{m}} \\
  {\text{D}}.\;50\left( {\sqrt 3 + 1} \right){\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 h be the height of the light house, which is represented by CD. Let E be the position of ship at $45^o$ and F be the position of ship at $30^o$. According to the question, EF = 100m. Now we will apply trigonometric formulas in triangles DCE and DCF.

$

  In\vartriangle DCE, \\

  \tan {45^o} = \dfrac{{DC}}{{CE}} \\

  1 = \dfrac{h}{{CE}} \\

$

h = CE …(1)

$

  In\;\vartriangle DCF, \\

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

  \dfrac{1}{{\sqrt 3 }} = \dfrac{{\text{h}}}{{CF}} \\

  By\;figure,\;CF = CE + EF \\

  {\text{h}} = \dfrac{{CE + EF}}{{\sqrt 3 }} = \dfrac{{CE + 100}}{{\sqrt 3 }} \\

$

Using equation (1) we can write that-

$

  {\text{h}} = \dfrac{{{\text{h}} + 100}}{{\sqrt 3 }} \\

  \sqrt 3 {\text{h}} - {\text{h}} = 100 \\

  {\text{h}} = \dfrac{{100}}{{\sqrt 3 - 1}} \\

$
Hence, the correct answer 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.