Two poles standing on a horizontal ground are of height x meters and 40 meters respectively. The line joining their tops makes an angle of `30^(@)` with the ground and the distance between the foot of the poles is `30sqrt3` meters, then the value of x can be

Options

A. 20
B. 30
C. 10
D. 50

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry of the Problem
We have two poles of heights \( x \) meters and \( 40 \) meters, respectively. The distance between the bases of the poles is \( 30\sqrt{3} \) meters. The line joining the tops of the poles makes an angle of \( 30^\circ \) with the ground.

### Step 2: Set Up the Diagram
- Let the foot of the pole of height \( x \) be point A and the foot of the pole of height \( 40 \) meters be point B.
- The height of pole A is \( x \) and the height of pole B is \( 40 \).
- The distance between points A and B is \( 30\sqrt{3} \) meters.

### Step 3: Identify the Angles
The angle made by the line joining the tops of the poles (let's call it line AB) with the ground is \( 30^\circ \).

### Step 4: Use the Tangent Function
From the geometry, we can use the tangent of the angle to relate the heights of the poles and the distance between their bases:
\[ \tan(30^\circ) = \frac{\text{Height difference}}{\text{Distance between the bases}} \]
The height difference between the two poles is \( 40 - x \) meters, and the distance between the bases is \( 30\sqrt{3} \) meters.

### Step 5: Substitute Values
Using the value of \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \):
\[ \frac{1}{\sqrt{3}} = \frac{40 - x}{30\sqrt{3}} \]

### Step 6: Cross-Multiply
Cross-multiplying gives:
\[ 30\sqrt{3} = \sqrt{3}(40 - x) \]

### Step 7: Simplify
Dividing both sides by \( \sqrt{3} \) (since \( \sqrt{3} \neq 0 \)):
\[ 30 = 40 - x \]

### Step 8: Solve for \( x \)
Rearranging the equation:
\[ x = 40 - 30 \]
\[ x = 10 \]

### Conclusion
Thus, the value of \( x \) is \( 10 \) meters.