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The problem asks to calculate the angle \( \alpha \) between two lines in the depicted cube to an accuracy of \( 1^\circ \). It mentions that you can solve the problem by two methods: by analyzing triangles or by calculating the angle between vectors.

### Steps for solving this problem:
Let’s determine the lines involved and solve it using the vector method.

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### 1. **Understand the geometry of the cube:**
- The given figure shows a cube, and the angle \( \alpha \) is between the diagonal \( A_1C_1 \) on the top face and the diagonal \( A_1B \) that spans two adjacent faces.

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### 2. **Assign coordinates to the vertices of the cube:**
Assume the cube has side length \( a = 1 \), and it is aligned with the coordinate axes. The vertices have the following coordinates:
- \( A_1 = (0, 0, 1) \)
- \( B_1 = (1, 0, 1) \)
- \( C_1 = (1, 1, 1) \)
- \( A = (0, 0, 0) \)
- \( B = (1, 0, 0) \)
- \( C = (1, 1, 0) \)

---

### 3. **Define the two vectors:**
- \( \vec{A_1C_1} = C_1 - A_1 = (1, 1, 0) - (0, 0, 1) = (1, 1, -1) \)
- \( \vec{A_1B} = B - A_1 = (1, 0, 0) - (0, 0, 1) = (1, 0, -1) \)

---

### 4. **Calculate the angle between the two vectors:**
The angle \( \alpha \) between two vectors is given by the formula:
\[
\cos \alpha = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}
\]
Substitute the vectors \( \vec{u} = \vec{A_1C_1} = (1, 1, -1) \) and \( \vec{v} = \vec{A_1B} = (1, 0, -1) \).

#### Dot product:
\[
\vec{A_1C_1} \cdot \vec{A_1B} = (1)(1) + (1)(0) + (-1)(-1) = 1 + 0 + 1 = 2
\]

#### Magnitudes:
\[
|\vec{A_1C_1}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}
\]
\[
|\vec{A_1B}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}
\]

#### Cosine of the angle:
\[
\cos \alpha = \frac{\vec{A_1C_1} \cdot \vec{A_1B}}{|\vec{A_1C_1}| |\vec{A_1B}|} = \frac{2}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}
\]

---

### 5. **Calculate the angle \( \alpha \):**
\[
\alpha = \arccos\left(\frac{\sqrt{6}}{3}\right)
\]
Using a calculator or approximate values:
\[
\alpha \approx 35.26^\circ
\]

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### Final Answer:
The angle \( \alpha \) between the lines is approximately **35°** (to the nearest degree).

Would you like a step-by-step explanation of the triangle method or more details about any of the steps above?

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### 5 Related Questions:
1. How is the dot product of two vectors calculated geometrically and algebraically?
2. What is the relationship between angles and vector components in three dimensions?
3. Can you calculate angles between planes using a similar vector-based method?
4. How does the choice of coordinate system affect the calculation of angles in geometry?
5. Why is it important to normalize vector magnitudes when calculating angles?

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### Tip:
When solving geometry problems involving angles in 3D space, always visualize the vectors or lines using coordinates to simplify the problem.

Brežinyje pavaizduotas kubas. Apskaičiuok pažymėto kampo α tarp tiesių didumą 1° tikslumu. (Uždavinį gali spręsti dviem būdais - nagrinėdamas trikampius arba ieškodamas kampo tarp vektorių.)

Learn how to calculate the angle between two lines in a cube with a step-by-step explanation. Using vector dot products and magnitudes, this solution explains how to find the angle to the nearest degree. Suitable for high school geometry and trigonometry students.

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