 The diagram shows a triangular prism ABCDQP of length 7cm� The cross-section is triangle PAB with PA = 4 cm, AB = 5 cm and angle PAB = 90°� Calculate the angle between the line PC and the base ABCD�

To find the angle between the line \( PC \) and the base \( ABCD \) of the triangular prism \( ABCDQP \), let's break down the problem step-by-step.

### Step 1: Understand the Geometry of the Prism
- The prism has a triangular cross-section \( PAB \) with \( PA = 4 \, \text{cm} \), \( AB = 5 \, \text{cm} \), and \( \angle PAB = 90^\circ \). This makes \( PAB \) a right-angled triangle.
- \( ABCD \) is a rectangle with sides \( AB \) and \( PQ = CD = 7 \, \text{cm} \), since it is the base of the prism.
- The points \( P \) and \( C \) are on opposite vertices along the height of the prism.

### Step 2: Position Vectors and Coordinates
- Place \( P \) at the origin \( (0, 0, 0) \).
- Point \( A \) is at \( (4, 0, 0) \), since \( PA = 4 \, \text{cm} \).
- Point \( B \) is at \( (4, 5, 0) \) because \( AB = 5 \, \text{cm} \) and \( \angle PAB = 90^\circ \).
- Point \( C \) is directly above \( B \) along the height of the prism, so \( C = (4, 5, 7) \).

### Step 3: Find the Direction Vector of Line \( PC \)
The direction vector from \( P \) to \( C \) is:
\[
\overrightarrow{PC} = (4 - 0, 5 - 0, 7 - 0) = (4, 5, 7)
\]

### Step 4: Find the Normal Vector of the Base \( ABCD \)
The base \( ABCD \) lies in the \( xy \)-plane, so its normal vector is along the \( z \)-axis:
\[
\text{Normal Vector to } ABCD = (0, 0, 1)
\]

### Step 5: Calculate the Angle Between \( PC \) and the Base \( ABCD \)
The angle \( \theta \) between line \( PC \) and the plane \( ABCD \) can be found using the dot product formula:

\[
\cos \theta = \frac{\overrightarrow{PC} \cdot \text{Normal Vector}}{|\overrightarrow{PC}| \cdot |\text{Normal Vector}|}
\]

Substituting in the values:

\[
\overrightarrow{PC} \cdot \text{Normal Vector} = (4, 5, 7) \cdot (0, 0, 1) = 7
\]

Magnitude of \( \overrightarrow{PC} \):

\[
|\overrightarrow{PC}| = \sqrt{4^2 + 5^2 + 7^2} = \sqrt{16 + 25 + 49} = \sqrt{90}
\]

Magnitude of the Normal Vector:

\[
|\text{Normal Vector}| = \sqrt{0^2 + 0^2 + 1^2} = 1
\]

Thus,

\[
\cos \theta = \frac{7}{\sqrt{90}}
\]

\[
\theta = \cos^{-1} \left(\frac{7}{\sqrt{90}}\right)
\]

Simplifying further:

\[
\theta \approx \cos^{-1} \left(\frac{7}{9.49}\right)
\]

\[
\theta \approx \cos^{-1} (0.737)
\]

\[
\theta \approx 42.8^\circ
\]

Therefore, the angle between the line \( PC \) and the base \( ABCD \) is approximately \( 42.8^\circ \).

Would you like more details or have any questions? Here are some related questions:

1. How do you calculate the volume of the prism?
2. What are the coordinates of point \( Q \)?
3. How would you find the angle between another line and the base?
4. How is the area of the base \( ABCD \) calculated?
5. What is the length of the diagonal of the base \( ABCD \)?

**Tip:** When dealing with angles between lines and planes, always remember that the dot product and magnitudes are your key tools for calculating these angles efficiently.

Learn how to calculate the angle between line PC and the base ABCD of a triangular prism with a detailed step-by-step solution. This problem covers geometry, trigonometry, vectors, and the dot product formula. Suitable for high school students.

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