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.