It is observed that on walking `x` meters towards a tower in a horizontal line through its base, the elevation of its top changes from `30^@` to `60^@`. The height of the tower is:

Options

A. `3sqrt2x`
B. `2sqrt3x`
C. `sqrt3/2x`
D. `2/sqrt3x`

Answer

To find the height of the tower based on the given angles of elevation, we can follow these steps:

### Step 1: Draw a diagram
Draw a vertical line representing the tower (AB) and a horizontal line representing the ground. Mark point C as the original position where the angle of elevation is 30° and point D as the position after walking x meters towards the tower where the angle of elevation is 60°.

### Step 2: Identify triangles
In the diagram, we have two right triangles:
1. Triangle ACB (where angle ACB = 30°)
2. Triangle ADB (where angle ADB = 60°)

### Step 3: Use trigonometric ratios
For triangle ACB (with angle 30°):
\[ \tan(30°) = \frac{AB}{BC} \]
Using the value of \(\tan(30°) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{AB}{BC} \implies AB = \frac{BC}{\sqrt{3}} \tag{1} \]

For triangle ADB (with angle 60°):
\[ \tan(60°) = \frac{AB}{BD} \]
Using the value of \(\tan(60°) = \sqrt{3}\):
\[ \sqrt{3} = \frac{AB}{BD} \implies AB = BD \cdot \sqrt{3} \tag{2} \]

### Step 4: Relate BD and BC
Since BD is the distance from point D to point B, we can express it as:
\[ BD = BC - x \]
Substituting this into equation (2):
\[ AB = (BC - x) \cdot \sqrt{3} \tag{3} \]

### Step 5: Substitute equation (1) into equation (3)
From equation (1), we can express BC in terms of AB:
\[ BC = AB \cdot \sqrt{3} \]
Substituting this into equation (3):
\[ AB = (AB \cdot \sqrt{3} - x) \cdot \sqrt{3} \]
Expanding this gives:
\[ AB = AB \cdot 3 - x \cdot \sqrt{3} \]

### Step 6: Rearranging the equation
Rearranging the equation to isolate AB:
\[ AB - 3AB = -x \cdot \sqrt{3} \]
\[ -2AB = -x \cdot \sqrt{3} \]
\[ 2AB = x \cdot \sqrt{3} \]
\[ AB = \frac{x \cdot \sqrt{3}}{2} \]

### Conclusion
The height of the tower (AB) is:
\[ AB = \frac{\sqrt{3}}{2} x \]