Each side of a square subtends an angle of `60^(@)` at the top of a tower 5 meters high standing at the center of the square. If a meters is the length of each side of the square, then a is equal to (use `sqrt2=1.41`)

Answer

To solve the problem, we need to find the length of each side of the square (denoted as \( a \)) given that each side subtends an angle of \( 60^\circ \) at the top of a tower that is 5 meters high.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have a tower of height \( h = 5 \) meters at the center of a square.
- Each side of the square subtends an angle of \( 60^\circ \) at the top of the tower.

2. **Setting Up the Triangle**:
- Let \( A \) and \( B \) be the endpoints of one side of the square, and \( C \) be the top of the tower.
- The angle \( \angle ACB = 60^\circ \).

3. **Using Properties of Isosceles Triangle**:
- Since \( AC = BC \) (the distances from the top of the tower to the endpoints of the side of the square are equal), triangle \( ACB \) is isosceles.
- The remaining angles \( \angle CAB \) and \( \angle CBA \) are each \( 60^\circ \), making triangle \( ACB \) equilateral.

4. **Finding the Length of Side \( AB \)**:
- Let \( D \) be the midpoint of \( AB \). The length \( AD = \frac{a}{2} \).
- In triangle \( ACD \), we have:
- \( CD = h = 5 \) meters (height of the tower).
- \( AD = \frac{a}{2} \).

5. **Using Trigonometric Ratios**:
- In triangle \( ACD \), we can use the tangent function:
\[ \tan(30^\circ) = \frac{CD}{AD} \]
- Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{5}{\frac{a}{2}} \]
- Cross-multiplying gives:
\[ 5 \cdot \sqrt{3} = \frac{a}{2} \]
- Therefore, \( a = 10 \sqrt{3} \).

6. **Finding the Length of Each Side**:
- Now, substituting \( \sqrt{3} \approx 1.41 \):
\[ a = 10 \cdot 1.41 = 14.1 \text{ meters} \]

Thus, the length of each side of the square is approximately \( 14.1 \) meters.