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To solve this problem, we need to find the angle between the line \( QT \) and the plane \( PQRS \) of the rectangular-based pyramid \( PQRST \). 

### Step-by-Step Solution:

1. **Understand the Geometry of the Pyramid**:
   - \( PQRS \) is the base of the pyramid, and it is a rectangle.
   - The height of the pyramid \( T \) is 24 cm above the center of the rectangle \( PQRS \).
   - The dimensions of the rectangle are \( PQ = 22 \) cm and \( PS = 25 \) cm.

2. **Find the Position of \( Q \)**:
   - \( Q \) is the midpoint of \( PR \). To find \( Q \), we first need the coordinates of \( P \) and \( R \).
   - Assume \( P \) is at the origin \( (0, 0) \), \( Q \) will be at \( \left(\frac{22}{2}, 0\right) = (11, 0) \).

3. **Coordinates of Points**:
   - \( P(0, 0, 0) \)
   - \( Q(11, 0, 0) \)
   - \( R(22, 0, 0) \)
   - \( S(0, 25, 0) \)
   - The point \( T \), being directly above the center of the base rectangle, has coordinates \( T(11, 12.5, 24) \).

4. **Find the Line \( QT \)**:
   - \( QT \) connects the point \( Q(11, 0, 0) \) to \( T(11, 12.5, 24) \).
   - The direction vector of \( QT \) is \( T - Q = (11 - 11, 12.5 - 0, 24 - 0) = (0, 12.5, 24) \).

5. **Normal to Plane \( PQRS \)**:
   - The plane \( PQRS \) is horizontal, so its normal vector is along the \( z \)-axis, which is \( (0, 0, 1) \).

6. **Find the Cosine of the Angle**:
   - The angle \( \theta \) between line \( QT \) and the plane \( PQRS \) can be found using the dot product:
   \[
   \cos \theta = \frac{\text{Vector } QT \cdot \text{Normal to } PQRS}{\|QT\| \cdot \|Normal\|}
   \]
   - The dot product \( (0, 12.5, 24) \cdot (0, 0, 1) = 24 \).
   - Magnitude of \( QT \), \( \|QT\| = \sqrt{0^2 + 12.5^2 + 24^2} = \sqrt{0 + 156.25 + 576} = \sqrt{732.25} \).
   - Magnitude of the normal vector \( \|Normal\| = 1 \).

   \[
   \cos \theta = \frac{24}{\sqrt{732.25}}
   \]

7. **Calculate the Angle**:
   \[
   \cos \theta = \frac{24}{\sqrt{732.25}}
   \]

   Let's calculate the angle \( \theta \) in degrees:

\[
\theta = \cos^{-1}\left(\frac{24}{\sqrt{732.25}}\right)
\]

Let's compute this.The angle between the line \(QT\) and the plane \(PQRS\) is approximately **27.5 degrees** when rounded to one decimal place.

Would you like more details on this calculation or have any questions? 

Here are some related questions you might find interesting:

1. How do you calculate the dot product between two vectors?
2. What is the significance of the normal vector to a plane?
3. How can you find the height of a pyramid given the dimensions of its base and its slant height?
4. What are some other methods to find the angle between a line and a plane?
5. How can you determine if a line is perpendicular or parallel to a plane?

**Tip:** Always double-check your geometric interpretations, especially when working with 3D objects, to ensure the correct coordinates and vectors are used.

Find the angle between the line QT and the plane PQRS of a rectangular-based pyramid PQRST, where the pyramid has a height of 24 cm, the base PQRS is a rectangle with PQ = 22 cm and PS = 25 cm.

Learn how to find the angle between the line QT and the plane PQRS in a rectangular-based pyramid PQRST. This problem involves geometric interpretation, vector calculations, and trigonometric concepts. Suitable for high school students in Grades 10-12.

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