A vertical pole PO is standing at the centre O of a square ABCD. If AC subtends an angle of `90^@` at the top , P of the pole, then the angle subtended by a side of the square at P is

Options

A. `45^@`
B. `30^@`
C. `60^@`
D. none of these

Explanation

To solve the problem step by step, let's break it down clearly.

**Given:**
- A vertical pole PO is standing at the center O of a square ABCD.
- The diagonal AC subtends an angle of 90° at the top P of the pole.

**To Find:**
- The angle subtended by a side of the square at P.

### Step-by-Step Solution:

1. **Understanding the Geometry:**
- Let the side length of the square ABCD be \( A \).
- The center O of the square divides the square into four equal parts.
- The diagonal AC of the square will have a length of \( AC = \sqrt{2}A \) (since the diagonal of a square is given by \( \sqrt{2} \times \text{side} \)).

**Hint:** Remember the formula for the diagonal of a square.

2. **Positioning the Points:**
- Place the square ABCD in a coordinate system:
- A = \((-A/2, A/2)\)
- B = \((A/2, A/2)\)
- C = \((A/2, -A/2)\)
- D = \((-A/2, -A/2)\)
- O = \((0, 0)\) (the center of the square)
- P = \((0, 0, h)\) (the top of the pole at height \( h \))

**Hint:** Visualize the square in a 3D coordinate system to understand the positions of points.

3. **Using the Given Angle:**
- The angle subtended by the diagonal AC at point P is given as 90°.
- This means that the lines PA and PC create a right triangle with the diagonal AC.

**Hint:** Recall that in a right triangle, the angles opposite the equal sides are equal.

4. **Finding the Angles at P:**
- Since the angle subtended by AC at P is 90°, the angles \( \angle AOP \) and \( \angle COP \) are both 45° (because the diagonal bisects the angle at the center).

**Hint:** Use the property of angles in an isosceles triangle.

5. **Finding the Lengths:**
- The distance OA (from O to A) is \( \frac{A}{\sqrt{2}} \) (half the diagonal).
- The height of the pole is \( h \), so we can use the Pythagorean theorem in triangle AOP:
\[ AP^2 = AO^2 + OP^2 \]
\[ AP^2 = \left(\frac{A}{\sqrt{2}}\right)^2 + h^2 \]
\[ AP^2 = \frac{A^2}{2} + h^2 \]

**Hint:** Use the Pythagorean theorem to relate the sides of the triangle.

6. **Considering the Side of the Square:**
- Now, consider the triangle ABP, where AB is a side of the square.
- Since OA = OB, the angle subtended by side AB at P will also be equal to the angle subtended by AC.

**Hint:** Use symmetry in the square to relate angles.

7. **Calculating the Angle:**
- Since triangle APB is isosceles (AP = PB), we can find the angle \( \alpha \) subtended by side AB at P.
- The angles at P will be equal, and since the total angle around point P is 360°, we can deduce:
\[ \alpha + \alpha + 90° = 180° \]
\[ 2\alpha = 90° \]
\[ \alpha = 45° \]

**Hint:** Use the sum of angles in a triangle to find the unknown angle.

8. **Conclusion:**
- Therefore, the angle subtended by a side of the square at point P is \( 45° \).

### Final Answer:
The angle subtended by a side of the square at P is \( 45° \).