The angle of elevation of a stationary cloud from a point 25 m above a lake is `30^(@)` and the angle of depression of its reflection in the lake is `60^(@)` . What is the height of the cloud above that lake-level ?

Answer

To find the height of the cloud above the lake level, we can follow these steps:

### Step 1: Understand the Problem
We have a point \( P \) which is 25 m above the lake. From this point, the angle of elevation to the cloud \( C \) is \( 30^\circ \), and the angle of depression to the reflection of the cloud \( D \) in the lake is \( 60^\circ \).

### Step 2: Draw the Diagram
Draw a vertical line representing the lake. Mark point \( P \) 25 m above the lake. Draw the cloud \( C \) above the lake and its reflection \( D \) below the lake. The angles of elevation and depression can be represented accordingly.

### Step 3: Define Variables
Let \( h \) be the height of the cloud \( C \) above the lake. The distance from point \( P \) to the cloud \( C \) is represented by the vertical segment \( PC \), and the distance from point \( P \) to the reflection \( D \) is represented by the vertical segment \( PD \).

### Step 4: Use Trigonometry for Cloud \( C \)
In triangle \( POC \):
- The angle of elevation \( \angle CPO = 30^\circ \).
- Using the tangent function:
\[ \tan(30^\circ) = \frac{CO}{PO} \]
Let \( CO = h \) (height of the cloud above the lake) and \( PO = d \) (horizontal distance from point \( P \) to the point directly below the cloud). Thus:
\[ \tan(30^\circ) = \frac{h}{d} \]
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):
\[ \frac{1}{\sqrt{3}} = \frac{h}{d} \implies d = h \sqrt{3} \quad \text{(Equation 1)} \]

### Step 5: Use Trigonometry for Reflection \( D \)
In triangle \( POD \):
- The angle of depression \( \angle DPO = 60^\circ \).
- The height \( PD \) can be expressed as \( PD = 25 + h \) (25 m above the lake plus the height of the cloud).
Using the tangent function:
\[ \tan(60^\circ) = \frac{OD}{PO} \]
Thus:
\[ \tan(60^\circ) = \frac{25 + h}{d} \]
Since \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{25 + h}{d} \implies d = \frac{25 + h}{\sqrt{3}} \quad \text{(Equation 2)} \]

### Step 6: Equate the Two Equations
From Equation 1 and Equation 2, we have:
\[ h \sqrt{3} = \frac{25 + h}{\sqrt{3}} \]
Cross-multiplying gives:
\[ h \cdot 3 = 25 + h \]
This simplifies to:
\[ 3h - h = 25 \implies 2h = 25 \implies h = \frac{25}{2} = 12.5 \text{ m} \]

### Step 7: Find the Total Height of the Cloud Above the Lake
The total height of the cloud above the lake level is:
\[ \text{Height of cloud} = h + 25 = 12.5 + 25 = 37.5 \text{ m} \]

### Final Answer
The height of the cloud above the lake level is **37.5 m**.

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