Let's break down the problem and solve it step-by-step.

Problem Recap:

We are given a cuboid $ABCDEFGH$, with:

• $BC = 6.8 \, \text{cm}$
• $DE = 9.3 \, \text{cm}$
• $\angle BAC = 52^\circ$

We are asked to find the size of the angle between the line $CF$ and the plane $ABCD$. Let's analyze this in parts.

Key Geometrical Elements:

1. Cuboid Shape:

   • $ABCD$ is the base of the cuboid.
   • $CF$ is a diagonal of one of the vertical faces $CDFH$.

2. Angle Interpretation:

   • We are tasked to find the angle between the diagonal $CF$ and the base plane $ABCD$. This is the angle between a diagonal line and a horizontal plane.

Step-by-Step Approach:

1. Coordinates Setup: Let's assign coordinates to the vertices of the cuboid:

   • $A = (0, 0, 0)$
   • $B = (6.8, 0, 0)$ [since $BC = 6.8 \, \text{cm}$]
   • $C = (6.8, 9.3, 0)$ [since $DE = 9.3 \, \text{cm}$ and $AB \parallel DC$]
   • $F = (6.8, 9.3, h)$, where $h$ is the height of the cuboid (unknown for now).

2. Find the Height $h$: Using the given angle $\angle BAC = 52^\circ$, we can find the height of the cuboid:

   • In the triangle $BAC$, we know:

     • $AB = BC = 6.8 \, \text{cm}$
     • $AC = \sqrt{AB^2 + BC^2} = \sqrt{(6.8)^2 + (9.3)^2} = \sqrt{46.24 + 86.49} = \sqrt{132.73} \approx 11.52 \, \text{cm}$

   • We also have: $\tan(\angle BAC) = \frac{\text{height } (h)}{\text{distance } (AC)}$ $\tan(52^\circ) = \frac{h}{11.52}$ $h = 11.52 \times \tan(52^\circ) \approx 11.52 \times 1.2799 \approx 14.74 \, \text{cm}$

3. Calculate the Angle: The next step is to compute the angle between the diagonal $CF$ and the base plane $ABCD$.

   • The vector $\overrightarrow{CF} = (0, 0, h)$ because the diagonal is only in the vertical direction.
   • The plane $ABCD$ is flat, so the angle we need is the arctangent of the height $h$ over the diagonal's horizontal projection (which is $BC = 6.8 \, \text{cm}$).

   Thus, we have: $\tan(\theta) = \frac{h}{BC}$ $\tan(\theta) = \frac{14.74}{6.8}$ $\theta = \tan^{-1} \left( \frac{14.74}{6.8} \right) \approx \tan^{-1}(2.167)$ $\theta \approx 65.9^\circ$

Final Answer:

The angle between the line $CF$ and the plane $ABCD$ is approximately 65.9°.

Let me know if you'd like further details or have any questions!

Here are 5 related questions to expand your understanding:

1. How would the problem change if the height of the cuboid were halved?
2. What if the angle $\angle BAC$ were $60^\circ$? How would that affect the height calculation?
3. How can we calculate the diagonal of the cuboid instead of just the face diagonal?
4. Can you find the angle between $CF$ and the vertical edge $CD$?
5. How would this solution differ if the cuboid were tilted or rotated?

Tip: Always visualize the problem by sketching out the 3D geometry when solving for angles in space!