Question

The angle of elevation of a cloud $C$ from a point $P , 200 m$ above a still lake is $30^{\circ} .$ If the angle of depression of the image of $C$ in the lake from the point $P$ is $60^{\circ}$, then $P C$ (in $m$ ) is equal to :

• A. 400
• B. $400 \sqrt{3}$
• C. 100
• D. $200 \sqrt{3}$

Answer 1

The Correct Option is A

To solve this problem, we need to visualize the scenario involving the cloud, its image in the lake, and the point 'P'. Consider the problem as a typical trigonometric heights and distances problem. Here is how you can solve it step-by-step:

1. Let 'C' be the actual position of the cloud, and 'C'' be its image in the lake. Since the image of the cloud is a reflection, the vertical distance between 'C' and the lake (let's denote this distance as 'h') is equal to the vertical distance between 'C'' and the lake, essentially making 'C'' exactly 2h below 'C'.
2. Point 'P' is given to be 200 m above the surface of the lake.
3. The angle of elevation from 'P' to the cloud 'C' is \(30^{\circ}\). The angle of depression from 'P' to the image of the cloud 'C'' is \(60^{\circ}\).
4. Triangles involved are:
   • Right triangle \( \triangle PCQ \), where Q is the point on the vertical line through the lake directly below cloud C.
   • Right triangle \( \triangle PC'Q' \), where Q' is the point on the vertical line through the lake directly below cloud C'.

Let's solve it by applying trigonometric ratios:

1. In \( \triangle PCQ \):
   1. Using angle of elevation \(30^{\circ}\), we have: \[ \tan 30^{\circ} = \frac{h}{x} = \frac{PC}{PQ} \] - where h is the height of the cloud above the lake and x is the horizontal distance from 'P' directly beneath 'C'.

   So, x = \frac{h}{\tan 30^{\circ}} = h \sqrt{3}.

2. In \( \triangle PC'Q' \):
   1. Using angle of depression \(60^{\circ}\): \[ \tan 60^{\circ} = \frac{2h + 200}{x} = \frac{PC'}{PQ'} \] - where x is again the horizontal distance from point 'P' to the image 'C'' vertically.

   So, x = \frac{2h + 200}{\tan 60^{\circ}} = \frac{2h + 200}{\sqrt{3}}.

Equating the two expressions for x, we get:

\[ h \sqrt{3} = \frac{2h + 200}{\sqrt{3}} \]

Multiply through by \sqrt{3} and solve:

1. \( 3h = 2h + 200 \)

Solving this gives: \begin{align*} 3h - 2h &= 200 \\ h &= 200 \end{align*}

Now, PC (distance from P to C) using hypotenuse formula: \begin{align*} PC = \frac{h}{\sin 30^{\circ}} = \frac{200}{0.5} = 400 \, m \end{align*}

The correct answer is therefore 400 meters.