Two men are on opposite sides of a tower .They measure the angles of elevation of the top of the tower as `30^(@)and45^(@)` respectively .If the height of the tower is 50 metre , the distance between the two men is (Take `sqrt(3)=1.73)`

Options

A. 136.5 metre
B. `50sqrt(3)` metre
C. `100sqrt(3)` metre
D. `135 .5` metre

Answer

To solve the problem, we will use trigonometric ratios to find the distances from each man to the base of the tower.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the height of the tower be \( h = 50 \) meters.
- Let \( A \) be the position of the first man, who measures an angle of elevation of \( 30^\circ \).
- Let \( B \) be the position of the second man, who measures an angle of elevation of \( 45^\circ \).
- Let \( C \) be the base of the tower.

2. **Using Trigonometry for Man A**:
- For man A, using the tangent function:
\[ \tan(30^\circ) = \frac{h}{AC} \]
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).
- Therefore:
\[ \frac{1}{\sqrt{3}} = \frac{50}{AC} \]
- Rearranging gives:
\[ AC = 50 \sqrt{3} \]

3. **Using Trigonometry for Man B**:
- For man B, using the tangent function:
\[ \tan(45^\circ) = \frac{h}{BC} \]
- We know that \( \tan(45^\circ) = 1 \).
- Therefore:
\[ 1 = \frac{50}{BC} \]
- Rearranging gives:
\[ BC = 50 \]

4. **Calculating the Total Distance**:
- The total distance \( AB \) between the two men is:
\[ AB = AC + BC = 50\sqrt{3} + 50 \]

5. **Substituting the Value of \(\sqrt{3}\)**:
- Given \( \sqrt{3} = 1.73 \):
\[ AB = 50 \times 1.73 + 50 \]
- Calculating:
\[ AB = 86.5 + 50 = 136.5 \text{ meters} \]

### Final Answer:
The distance between the two men is **136.5 meters**.