They are given in mm. The ground line is parallel and the small paper below the axis is 2cm. Stump 0 from the left of the 4cm paper. They give A (0.70,20) B (60, -10.60).
1. Draw a right triangle ABC ABC Top C Horizontal e (x = 60, y = 90)
2. Find the orthogonal Ω of triangle ABC and the vertex D of the rectangle ABCD (D.ABC) which is a right triangle.
3. Find the cross section of the ABCD rectangle, the plan is parallel to the plan B1, the seal is 45.

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

### 1. **Draw a Right Triangle ABC**

Given:
- Points \( A(0.70, 20) \) and \( B(60, -10.60) \)
- The top vertex \( C \) is horizontally aligned at \( C(60, 90) \).

We can now determine the structure of triangle ABC.

1. **Vertices of Triangle ABC**:
   - \( A(0.70, 20) \)
   - \( B(60, -10.60) \)
   - \( C(60, 90) \)

This triangle is right-angled at vertex \( B \), as \( C \) is vertically above \( B \) (same x-coordinate).

#### Diagram:
You can visualize this triangle, where:
- Line AB represents one side, 
- Line BC is the vertical height from point \( B \) to \( C \),
- Line AC is the hypotenuse.

### 2. **Find the Orthogonal Ω and Vertex D**

#### Orthogonal Ω:
The orthogonal \( \Omega \) of triangle ABC refers to the line that is perpendicular to the plane of triangle ABC.

- Since we know that \( \overrightarrow{BC} \) is vertical, we can focus on the orthogonal that is perpendicular to the plane defined by the points \( A, B, C \).
- The orthogonal line passes through a point in space perpendicular to the surface.

#### Finding Vertex D:
- \( D \) forms a rectangle ABCD. Since triangle ABC is right-angled at \( B \), we need to find a point \( D \) such that \( ABCD \) is a rectangle, and \( D \) is positioned such that \( D \) makes a right angle with lines \( AB \) and \( BC \).

We know:
- \( AB \) is diagonal along the base.
- \( BC \) is vertical.

Thus, \( D \) is located horizontally from point \( A \) and vertically aligned with \( C \).

### 3. **Find the Cross Section of the Rectangle**

- The cross section refers to the intersection between the plane parallel to the rectangle and another plane.
- The plane \( B1 \) and a seal at 45° would define the angle for this cross-section.

Given that the plane B1 is parallel and the seal is inclined at 45°, the intersection of the ABCD rectangle with the seal will give you a section of the rectangle. You can compute the coordinates of the points of intersection by projecting along the angle of the seal.

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Would you like a more detailed breakdown of any of these points? Here are some related questions you could explore:

1. How do we calculate the lengths of sides of triangle ABC?
2. What is the area of triangle ABC?
3. How do we determine the slope of the hypotenuse \( AC \)?
4. How do we find the equation of the plane containing triangle ABC?
5. What does the 45° seal mean geometrically for the cross-section?

### Tip:  
Always start by analyzing the given points in relation to each other on the coordinate plane before solving more complex geometric properties.

They are given in mm. The ground line is parallel and the small paper below the axis is 2cm. Stump 0 from the left of the 4cm paper. They give A (0.70, 20) B (60, -10.60). 1. Draw a right triangle ABC ABC Top C Horizontal e (x = 60, y = 90) 2. Find the orthogonal Ω of triangle ABC and the vertex D of the rectangle ABCD (D.ABC) which is a right triangle. 3. Find the cross section of the ABCD rectangle, the plan is parallel to the plan B1, the seal is 45.

This detailed solution explores a geometry problem involving right triangles and cross sections. The problem involves points A (0.70, 20), B (60, -10.60), and C (60, 90), forming a right triangle ABC. Learn how to find the orthogonal Ω, vertex D, and the cross section of the rectangle ABCD. Suitable for students in Grades 10-12.

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