Question

At a point A 20 meters above the level of water in a lake, the angle of elevation of a cloud is ${{30}^{\circ }}$. The angle of depression of the reflection of the cloud in the lake at A is ${{60}^{\circ }}$. Find the distance of the cloud from point A.

Answer 1

Hint: In this question, the angle of elevation of the cloud from the point A is given to be ${{30}^{\circ }}$. Also, it is given that A is 20meters above the water level of the lake. However, in reflection, the depth of the image should be equal to the height of the cloud above the water level. Using this information, we can construct an appropriate figure and then use trigonometric ratios of the given angles to find the distance of the cloud from the point A.

Complete step-by-step answer:

It is given that point A is 20 meters above the water level of the lake (the line GQ is at the level of water of the lake). The angle of elevation of the cloud from point A is ${{30}^{\circ }}$. The angle of depression of the reflection of the cloud in the lake at A is ${{60}^{\circ }}$. Let us visualize the above information in the form of a helpful labelled diagram as shown in the figure.

Here, $\Delta ABC\text{ and }\Delta ABD$ are right angle triangles right angled at B and GQ is the lake water level.
$\begin{align}
  & \angle CAB={{30}^{\circ }}\text{ and} \\
 & \angle BAD={{60}^{\circ }} \\
 & AQ=QP=BG=GF=20mts..........(1.1) \\
\end{align}$
Let CB be x. So,
 $\begin{align}
  & BD=BG+GF+FD \\
 & \Rightarrow BD=20+20+x=(40+x)mts...........(1.2) \\
\end{align}$
(as D and P are the reflection of the cloud C and point A respectively)
$\begin{align}
  & In\text{ }\Delta ABC, \\
 & \tan {{30}^{\circ }}=\dfrac{CB}{AB} \\
 & \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{x}{AB} \\
 & \Rightarrow AB=\sqrt{3}x............(1.3) \\
\end{align}$
$\begin{align}
  & In\text{ }\Delta ABD, \\
 & \tan {{60}^{\circ }}=\dfrac{BD}{AB} \\
 & \Rightarrow \sqrt{3}=\dfrac{x+40}{AB} \\
 & \Rightarrow AB=\dfrac{x+40}{\sqrt{3}}.............(1.4) \\
\end{align}$
From equations 1.3 and 1.4, we get
$\begin{align}
  & AB=x\sqrt{3}=\dfrac{x+40}{\sqrt{3}} \\
 & \Rightarrow 3x=x+40 \\
 & \Rightarrow 3x-x=40 \\
 & \Rightarrow 2x=40 \\
 & \Rightarrow x=20mts \\
\end{align}$
So, $BC=20mts$.
Now, the distance between point A and the cloud is CA.
$\begin{align}
  & In\text{ }\Delta ABC, \\
 & \sin {{30}^{\circ }}=\dfrac{BC}{AC} \\
 & \Rightarrow \dfrac{1}{2}=\dfrac{20}{AC} \\
 & \Rightarrow AC=40mts \\
\end{align}$
Thus, we get our answer as 40 meters is the distance between the cloud and point A.

Note: We should note that while considering the point of reflection of the cloud in the water, it will be at the same depth as the cloud is over the level of the water of the lake. Also, we could have chosen any distance in the figure as x and solved the problem using trigonometric ratios as any one variable can be solved from the equations.