The angle of elevation of the top of a tower at a point G on the ground is `30^(@)`. On walking 20 m towards the tower the angle of elevation becomes `60^(@)`. The height of the tower is equal to :

Options

A. `10//sqrt3m`
B. `20sqrt3m`
C. `20//sqrt3m`
D. `10sqrt3m`

Answer

To solve the problem, we will use trigonometric principles involving right triangles formed by the tower and the angles of elevation.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let the height of the tower be \( H \).
- Let the distance from point G to the base of the tower be \( x \).
- After walking 20 m towards the tower, the new distance from the tower is \( x - 20 \).

2. **Set Up the First Triangle:**
- From point G, the angle of elevation to the top of the tower is \( 30^\circ \).
- Using the tangent function:
\[ \tan(30^\circ) = \frac{H}{x} \]
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{H}{x} \quad \Rightarrow \quad H = \frac{x}{\sqrt{3}} \quad \text{(Equation 1)} \]

3. **Set Up the Second Triangle:**
- After walking 20 m towards the tower, the angle of elevation is \( 60^\circ \).
- Using the tangent function again:
\[ \tan(60^\circ) = \frac{H}{x - 20} \]
- We know that \( \tan(60^\circ) = \sqrt{3} \), so:
\[ \sqrt{3} = \frac{H}{x - 20} \quad \Rightarrow \quad H = \sqrt{3}(x - 20) \quad \text{(Equation 2)} \]

4. **Equate the Two Expressions for H:**
- From Equation 1 and Equation 2, we have:
\[ \frac{x}{\sqrt{3}} = \sqrt{3}(x - 20) \]

5. **Clear the Fractions:**
- Multiply both sides by \( \sqrt{3} \):
\[ x = 3(x - 20) \]
- Simplifying this gives:
\[ x = 3x - 60 \]
- Rearranging:
\[ 60 = 3x - x \quad \Rightarrow \quad 60 = 2x \quad \Rightarrow \quad x = 30 \]

6. **Find the Height of the Tower:**
- Substitute \( x = 30 \) back into Equation 1 to find \( H \):
\[ H = \frac{30}{\sqrt{3}} = 10\sqrt{3} \]

### Conclusion:
The height of the tower is \( H = 10\sqrt{3} \) meters.