A vertical pole AB is standing at the centre B of a square PQRS . If PR subtands an angle of `90^(@)` at the top , A of the pole , then the angle subtended by a side of the square at A is :

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. None of these

Answer

To solve the problem step by step, we will analyze the given information and apply trigonometric principles.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a square PQRS with a vertical pole AB standing at the center B of the square.
- The diagonal PR of the square subtends an angle of 90 degrees at the top of the pole (point A).

2. **Identifying Key Points**:
- Let the side length of the square be \( x \).
- The center B divides the diagonal PR into two equal parts. The length of the diagonal PR can be calculated as \( PR = x\sqrt{2} \).

3. **Finding Lengths**:
- Since B is the center, the length of PB (or BR) is half of PR:
\[ PB = \frac{PR}{2} = \frac{x\sqrt{2}}{2} = \frac{x}{\sqrt{2}}. \]

4. **Analyzing Triangle PAB**:
- In triangle PAB, we know:
- \( \angle PAR = 90^\circ \) (given),
- \( PA = AR \) (since the pole is at the center of the square).
- Therefore, angles \( \angle APB \) and \( \angle ARB \) are both \( 45^\circ \) because the triangle PAB is isosceles.

5. **Using Pythagorean Theorem**:
- In triangle PAB:
\[ AB^2 + PB^2 = PA^2. \]
- We know \( PB = \frac{x}{\sqrt{2}} \) and \( PA = x \):
\[ AB^2 + \left(\frac{x}{\sqrt{2}}\right)^2 = x^2. \]
- Simplifying:
\[ AB^2 + \frac{x^2}{2} = x^2 \implies AB^2 = x^2 - \frac{x^2}{2} = \frac{x^2}{2} \implies AB = \frac{x}{\sqrt{2}}. \]

6. **Finding the Angle Subtended by Side PQ**:
- Now, consider triangle APQ. Since \( PA = AQ = PQ = x \) (all sides of the square are equal), triangle APQ is equilateral.
- Therefore, each angle in triangle APQ is \( 60^\circ \).

7. **Conclusion**:
- The angle subtended by a side of the square at point A is \( 60^\circ \).

### Final Answer:
The angle subtended by a side of the square at A is \( 60^\circ \).