A vertical tower subtends an angle of `60^(@)` at a point on the same level as the foot of the tower. On moving 100 m further in line with the tower, it subtends an angle of `30^(@)` at the point. Then, the height of the tower, it subtends an angle of `30^(@)` at the point. Then, the height of the tower is

Options

A. `20sqrt3m`
B. `30sqrt3m`
C. `50sqrt3`
D. `10sqrt3m`

Answer

To find the height of the tower, we can use trigonometric principles based on the angles subtended by the tower at different points.

### Step-by-step Solution:

1. **Define the Variables:**
- Let the height of the tower be \( h \).
- Let the distance from the foot of the tower to the first observation point be \( x \).
- The observer moves 100 m further away, making the new distance \( x + 100 \).

2. **Using the First Angle (60 degrees):**
- At the first point, the angle of elevation is \( 60^\circ \).
- From trigonometry, we know:
\[ \tan(60^\circ) = \frac{h}{x} \]
- Since \( \tan(60^\circ) = \sqrt{3} \), we can write:
\[ \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \text{(Equation 1)} \]

3. **Using the Second Angle (30 degrees):**
- At the second point (100 m further), the angle of elevation is \( 30^\circ \).
- Again, using trigonometry:
\[ \tan(30^\circ) = \frac{h}{x + 100} \]
- Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we can write:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x + 100} \implies h = \frac{(x + 100)}{\sqrt{3}} \quad \text{(Equation 2)} \]

4. **Equating the Two Expressions for \( h \):**
- From Equation 1 and Equation 2, we have:
\[ x\sqrt{3} = \frac{x + 100}{\sqrt{3}} \]
- Cross-multiplying gives:
\[ 3x = x + 100 \]
- Rearranging the equation:
\[ 3x - x = 100 \implies 2x = 100 \implies x = 50 \]

5. **Finding the Height \( h \):**
- Substitute \( x = 50 \) back into Equation 1:
\[ h = 50\sqrt{3} \]

### Final Answer:
The height of the tower is \( 50\sqrt{3} \) meters.

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