https://r2.erweima.ai/mathbot/image/1724381969133wzfy6s1j.jpg

To solve the problem in point 6, I'll need to extract the relevant information from the image. Let me analyze it.The text in point 6 reads:

"Un coche recorre desde el origen hasta el lago \( A \), 200 km de distancia en la dirección Norte 30° Oeste y luego 300 km hacia el este. Determine la distancia y dirección del coche con respecto al origen."

Let's break down the problem step by step.

### Problem Breakdown:
1. **First Displacement:**
   - The car travels 200 km in the direction North 30° West.
   - This vector can be broken down into its components:
     - Northward (Y-axis) component: \( 200 \times \cos(30°) \)
     - Westward (X-axis) component: \( 200 \times \sin(30°) \)
   - Since West is negative on the X-axis, the X component is negative.

2. **Second Displacement:**
   - The car then travels 300 km directly east.
   - This is a straightforward vector in the positive X-axis direction: \( +300 \) km.

3. **Total Displacement:**
   - Combine the X and Y components from both displacements.

4. **Distance from Origin:**
   - Use the Pythagorean theorem to find the magnitude of the resultant vector.
   - \( D = \sqrt{(\text{Total } X)^2 + (\text{Total } Y)^2} \)

5. **Direction from Origin:**
   - Use the arctangent function to determine the direction angle relative to the positive X-axis (East direction).
   - \( \theta = \arctan\left(\frac{\text{Total } Y}{\text{Total } X}\right) \)

Now, let's compute these values:

1. **Northward (Y-axis) Component:**
   \[
   200 \times \cos(30°) = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3} \approx 173.21 \text{ km North}
   \]

2. **Westward (X-axis) Component:**
   \[
   200 \times \sin(30°) = 200 \times 0.5 = 100 \text{ km West (or -100 km)}
   \]

3. **Total X Component:**
   \[
   -100 + 300 = 200 \text{ km East}
   \]

4. **Total Y Component:**
   \[
   173.21 \text{ km North}
   \]

5. **Distance from Origin:**
   \[
   D = \sqrt{(200)^2 + (173.21)^2} \approx \sqrt{40000 + 30000} \approx \sqrt{70000} \approx 264.58 \text{ km}
   \]

6. **Direction from Origin:**
   \[
   \theta = \arctan\left(\frac{173.21}{200}\right) \approx \arctan(0.866) \approx 40.89° \text{ North of East}
   \]

### Final Answer:
- **Distance from Origin:** \( \approx 264.58 \) km
- **Direction:** \( \approx 40.89° \) North of East

Would you like further details, or do you have any questions about the steps? Here are some related questions to consider:

1. How would the problem change if the car traveled South 30° East instead?
2. What would be the car's total displacement if the second leg of the journey was westward?
3. Can you explain how vector addition works in this context?
4. How does the angle affect the resultant displacement in vector problems?
5. What would happen to the resultant distance if both components were in the same direction?

**Tip:** When solving vector problems, always break down vectors into their components along the X and Y axes, then combine these components using vector addition.

Learn how to calculate the distance and direction of a car from its origin using vector components and the Pythagorean theorem. This problem involves breaking down vectors, combining components, and using trigonometric functions. Suitable for high school students studying vectors.

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