From the top of a lamp post of height x metres, two objects on the ground on the sameside of it ( and in line with the foot of the lamp post ) are observed at angles of depression of `30^(@)` and `60^(@)`, respectively. The distance between the objects is `32 sqrt( 3) `m. The value of x is `:`

Options

A. 54
B. 48
C. 45
D. 36

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Problem
We have a lamp post of height \( x \) meters. From the top of the lamp post, two objects on the ground are observed at angles of depression of \( 30^\circ \) and \( 60^\circ \). The distance between the two objects is \( 32\sqrt{3} \) meters. We need to find the height \( x \).

### Step 2: Draw a Diagram
Draw a vertical line representing the lamp post. Label the top of the lamp post as point \( A \) and the foot of the lamp post as point \( B \). Let the two objects on the ground be points \( C \) and \( D \), with \( C \) being closer to the lamp post than \( D \).

### Step 3: Identify the Angles
From point \( A \) (the top of the lamp post):
- The angle of depression to point \( C \) is \( 60^\circ \).
- The angle of depression to point \( D \) is \( 30^\circ \).

### Step 4: Set Up the Right Triangles
Using the angles of depression, we can set up two right triangles:
1. Triangle \( ABD \) for angle \( 60^\circ \).
2. Triangle \( ABC \) for angle \( 30^\circ \).

### Step 5: Use Trigonometric Ratios
For triangle \( ABD \):
- The tangent of \( 60^\circ \) is given by:
\[ \tan(60^\circ) = \frac{x}{y} \]
where \( y \) is the distance from point \( B \) to point \( C \).
Since \( \tan(60^\circ) = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{x}{y} \implies y = \frac{x}{\sqrt{3}} \tag{1} \]

For triangle \( ABC \):
- The tangent of \( 30^\circ \) is given by:
\[ \tan(30^\circ) = \frac{x}{y + 32\sqrt{3}} \]
where \( y + 32\sqrt{3} \) is the distance from point \( B \) to point \( D \).
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{x}{y + 32\sqrt{3}} \implies y + 32\sqrt{3} = \sqrt{3}x \tag{2} \]

### Step 6: Substitute Equation (1) into Equation (2)
Substituting \( y = \frac{x}{\sqrt{3}} \) from equation (1) into equation (2):
\[ \frac{x}{\sqrt{3}} + 32\sqrt{3} = \sqrt{3}x \]

### Step 7: Solve for \( x \)
Rearranging gives:
\[ 32\sqrt{3} = \sqrt{3}x - \frac{x}{\sqrt{3}} \]
Multiplying through by \( \sqrt{3} \) to eliminate the fraction:
\[ 32 \cdot 3 = 3x - x \]
\[ 96 = 2x \]
\[ x = \frac{96}{2} = 48 \]

### Final Answer
The height of the lamp post \( x \) is \( 48 \) meters.

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