https://r2.erweima.ai/mathbot/image/172199945216110gqfbdm.png

Let's find the angle between \(GB\) and the plane \(ABCD\).

Given:
- \(ABCD\) is a square paper with side \(12 \, \text{cm}\).
- \(EF\) is a line parallel to \(AD\) and \(AE : EB = 1:2\).
- The paper is folded along \(EF\) such that \(\angle AEB = 80^\circ\).
- \(G\) is the center of \(ADFE\).
- \(J\) is the projection of \(G\) on the table.

First, we need to understand the positions of \(G\) and \(B\):

1. **Find the Coordinates of Points:**
   - Since \(A\) is at \((0,0)\), \(B\) at \((12,0)\), \(C\) at \((12,12)\), and \(D\) at \((0,12)\).
   - \(E\) divides \(AB\) in the ratio \(1:2\), so \(E\) is at \((4,0)\).
   - Since \(EF\) is parallel to \(AD\), \(F\) is at \((4,12)\).

2. **Find the Coordinates of \(G\):**
   - \(G\) is the centroid of \(ADFE\). The coordinates of \(G\) can be calculated as the average of the coordinates of \(A, D, F, E\).
   - Coordinates of \(A = (0, 0)\), \(D = (0, 12)\), \(F = (4, 12)\), and \(E = (4, 0)\).
   - Therefore, \(G\) is at:
     \[
     G = \left( \frac{0+0+4+4}{4}, \frac{0+12+12+0}{4} \right) = \left( 2, 6 \right)
     \]

3. **Find the Coordinates of \(B\):**
   - Coordinates of \(B = (12, 0)\).

4. **Vector \(GB\):**
   - The vector \(GB\) can be found by subtracting the coordinates of \(G\) from \(B\):
     \[
     GB = (12 - 2, 0 - 6) = (10, -6)
     \]

5. **Projection of \(G\) on \(ABCD\):**
   - Since \(G\) is the center and lies directly above the plane \(ABCD\), its projection \(J\) on the plane will have the same \(x\) and \(y\) coordinates, but the \(z\)-coordinate will be 0.
   - Therefore, the coordinates of \(J\) will be \((2, 6, 0)\).

6. **Calculate the Angle Between \(GB\) and the Plane \(ABCD\):**
   - The plane \(ABCD\) lies flat on the table, hence its normal vector is along the \(z\)-axis.
   - The vector \(GB\) in 3D can be represented as \(GB = (10, -6, 6)\) (since the paper is folded, \(G\) is elevated above the plane \(ABCD\)).
   - The normal vector to \(ABCD\) is \( \mathbf{n} = (0, 0, 1)\).

7. **Angle Calculation:**
   - Use the dot product formula:
     \[
     \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}
     \]
   - Here, \( \mathbf{a} = GB = (10, -6, 6) \) and \( \mathbf{b} = \mathbf{n} = (0, 0, 1)\).
   - The dot product \( \mathbf{a} \cdot \mathbf{b} = 10 \cdot 0 + (-6) \cdot 0 + 6 \cdot 1 = 6 \).
   - The magnitude of \( \mathbf{a} \) is \( \| \mathbf{a} \| = \sqrt{10^2 + (-6)^2 + 6^2} = \sqrt{100 + 36 + 36} = \sqrt{172} \).
   - The magnitude of \( \mathbf{b} \) is \( \| \mathbf{b} \| = 1 \).

   Therefore,
   \[
   \cos \theta = \frac{6}{\sqrt{172}} = \frac{6}{\sqrt{4 \cdot 43}} = \frac{6}{2 \sqrt{43}} = \frac{3}{\sqrt{43}}
   \]
   Hence,
   \[
   \theta = \cos^{-1}\left( \frac{3}{\sqrt{43}} \right)
   \]

Therefore, the angle between \(GB\) and the plane \(ABCD\) is \(\cos^{-1}\left( \frac{3}{\sqrt{43}} \right)\).

Would you like details or have any questions? Here are 5 questions for further exploration:
1. How to find the coordinates of the centroid of a polygon?
2. What is the dot product and how is it used to find angles between vectors?
3. How to calculate the magnitude of a vector in 3D?
4. How can we represent planes in 3D and find their normal vectors?
5. How to use trigonometric functions to find angles from cosine values?

**Tip:** When finding the centroid of a shape, remember to average the coordinates of all vertices.

Learn how to calculate the angle between vector GB and the plane ABCD using 3D geometry principles. This problem involves finding coordinates, vectors, and using trigonometric functions. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation