Question

If the angle of elevation of a cloud from a point $h$ metres above a lake be $\theta $ and the angle of depression of its reflection in the lake be $\phi $, prove that the distance of the cloud from the point of observation is $\dfrac{{2h\cos \phi }}{{\sin \left( {\phi - \theta } \right)}}$. Also find the horizontal distance of the cloud from the place of observation.

Answer 1

Hint: In this question form two equation based on the data given i.e. $x + h = y\tan \phi ,x - h = y\tan \theta $ and use basic conversion like $\left( {\tan \phi - \tan \theta } \right) = \dfrac{{\sin \left( {\phi - \theta } \right)}}{{\cos \phi \cos \theta }}$. Use this to find the horizontal distance of the cloud from the place of observation.

Complete Step-by-Step solution:
According to the question we have to find $PL$, we have
$
  x + h = y\tan \phi \\
  x - h = y\tan \theta \\
$
Now, subtract
$
  2h = y\left( {\tan \phi - \tan \theta } \right) = \dfrac{{y\sin \left( {\phi - \theta } \right)}}{{\cos \phi \cos \theta }} \\
  \therefore y = \dfrac{{2h\cos \theta \cos \phi }}{{\sin \left( {\phi - \theta } \right)}} \\
 $
Above gives the horizontal distance of the cloud from the point of observation.
Again $y = PL = \cos \theta $
$\therefore PL = \dfrac{{2h\cos \phi }}{{\sin \left( {\phi - \theta } \right)}}$

Note: It is always advisable to remember some basic conversions and the concept of angle of elevation that denotes the angle from the horizontal upward to an object while involving trigonometric questions as it helps save a lot of time.