Question

If the angle of elevation of a cloud from a point $P$ which is $25\, m$ above a lake be $30^\circ$ and the angle of depression of reflection of the cloud in the lake from $P$ be $60^\circ$, then the height of the cloud (in meters) from the surface of the lake is :

• A. 42
• B. 50
• C. 45
• D. 60

Answer 1

The Correct Option is B

To solve this problem, we need to understand the scenario and apply trigonometric concepts. The point \( P \) is 25 meters above the lake. The angle of elevation to the cloud is \( 30^\circ \), and the angle of depression to the reflection of the cloud in the lake is \( 60^\circ \).

Let's denote:

• \( AB \) as the distance from the lake surface to the cloud, which is what we want to find.
• \( CD \) as the distance from the point \( P \) to the cloud.
• \( BD \) as the horizontal distance from the point directly below \( P \) to the base of the cloud.
• \( C \) as the reflection of the cloud in the lake, making \( CD = CB \).

Using trigonometry in \(\triangle PBD\):

The angle of elevation of the cloud is \( 30^\circ \).

Using the tangent of \( 30^\circ \):

\[ \tan 30^\circ = \frac{AB}{BD} \]

\[ \frac{1}{\sqrt{3}} = \frac{AB}{BD} \]

\[ BD = AB \cdot \sqrt{3} \]

For the reflection, the angle of depression is \( 60^\circ \).

Using the tangent of \( 60^\circ \):

\[ \tan 60^\circ = \frac{2 \times AB + 50}{BD} \]

\[ \sqrt{3} = \frac{2 \times AB + 50}{AB \cdot \sqrt{3}} \]

Cross multiplying, we get:

\[ 3AB = 2 \times AB + 50 \]

\[ AB = 50 \]

Therefore, the height of the cloud from the surface of the lake is 50 meters.