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We are given a rectangular-based pyramid \( PQRST \), where the base \( PQRS \) is a square with side lengths of 22 cm, and the height of the pyramid from the apex \( T \) to the center of the base is 20 cm. We are tasked with finding the angle between the line \( ST \) and the plane \( PQRS \).

### Approach:

1. **Coordinates Setup**: 
   - Place the center of the square base \( O \) at the origin \((0, 0, 0)\).
   - Let the vertices \( P, Q, R, S \) of the square base be at:
     - \( P(-11, -11, 0) \)
     - \( Q(11, -11, 0) \)
     - \( R(11, 11, 0) \)
     - \( S(-11, 11, 0) \)
   - The apex \( T \) is directly above \( O \) with coordinates \( (0, 0, 20) \).

2. **Vector for Line \( ST \)**:
   The vector \( \overrightarrow{ST} \) can be calculated using the coordinates of \( S \) and \( T \).
   \[
   \overrightarrow{ST} = (0 - (-11), 0 - 11, 20 - 0) = (11, -11, 20)
   \]

3. **Normal to the Plane \( PQRS \)**:
   Since \( PQRS \) is in the \( xy \)-plane, the normal vector to the plane is simply in the \( z \)-direction, which is:
   \[
   \mathbf{n} = (0, 0, 1)
   \]

4. **Angle Between Line and Plane**:
   The angle \( \theta \) between the line and the plane can be found using the dot product formula between the line's direction vector \( \mathbf{v} = \overrightarrow{ST} \) and the normal vector \( \mathbf{n} \):
   \[
   \cos \theta = \frac{|\mathbf{v} \cdot \mathbf{n}|}{|\mathbf{v}| |\mathbf{n}|}
   \]
   where:
   - \( \mathbf{v} = (11, -11, 20) \)
   - \( \mathbf{n} = (0, 0, 1) \)

   The dot product is:
   \[
   \mathbf{v} \cdot \mathbf{n} = (11)(0) + (-11)(0) + (20)(1) = 20
   \]

   The magnitude of \( \mathbf{v} \) is:
   \[
   |\mathbf{v}| = \sqrt{11^2 + (-11)^2 + 20^2} = \sqrt{121 + 121 + 400} = \sqrt{642} \approx 25.33
   \]

   The magnitude of \( \mathbf{n} \) is:
   \[
   |\mathbf{n}| = \sqrt{0^2 + 0^2 + 1^2} = 1
   \]

   So, we have:
   \[
   \cos \theta = \frac{|20|}{25.33 \times 1} \approx \frac{20}{25.33} \approx 0.7895
   \]

   Therefore, the angle \( \theta \) is:
   \[
   \theta = \cos^{-1}(0.7895) \approx 37.8^\circ
   \]

### Final Answer:
The angle between the line \( ST \) and the plane \( PQRS \) is approximately \( 37.8^\circ \).

Would you like more details on any of the steps?

### Relative Questions:
1. How do you find the direction vector of a line given two points?
2. What is the geometric interpretation of the dot product in vector math?
3. How do you find the normal vector of a plane in 3D space?
4. What is the physical meaning of an angle between a line and a plane?
5. Can you calculate angles between two lines in space using similar methods?

### Tip:
When dealing with geometric figures in 3D, assigning coordinates to points and using vector operations like dot products and cross products can simplify finding angles and distances.

Find the angle between the line ST and the plane PQRS in a rectangular-based pyramid with a height of 20 cm and a square base side length of 22 cm.

Learn how to find the angle between the line ST and the plane PQRS in a rectangular-based pyramid with a square base of 22 cm and height of 20 cm. This solution covers 3D geometry concepts, vector calculations, and the dot product formula, making it suitable for students in Grades 10-12.

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