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The image shows a vector problem in which the vector \( \vec{d} \) has a magnitude of 300 m and a direction of 40° north of east. The task is to break this vector into its horizontal (\( d_x \)) and vertical (\( d_y \)) components.

Given:
- Magnitude of the vector, \( d = 300 \, \text{m} \)
- Angle \( \theta = 40^\circ \)

To find the components:
- The horizontal component \( d_x = d \cdot \cos(\theta) \)
- The vertical component \( d_y = d \cdot \sin(\theta) \)

Let's calculate those values.The horizontal and vertical components of the vector are:

- \( d_x = 229.81 \, \text{m} \) (horizontal component)
- \( d_y = 192.84 \, \text{m} \) (vertical component)

Would you like further explanation or details on this process? Here are a few related questions for deeper exploration:

1. How do you resolve vectors using trigonometry?
2. What would happen if the angle were measured counterclockwise from north instead of east?
3. How does changing the magnitude affect the vector components?
4. What are the practical applications of resolving vectors into components?
5. How would you combine multiple vectors to find a resultant vector?

**Tip:** When dealing with angles in vector components, ensure you're measuring from the correct axis (east, north, etc.) to avoid mistakes!

A vector with a magnitude of 300 m is directed 40° north of east. How do you find the horizontal and vertical components of this vector?

Learn how to resolve a 300 m vector directed at 40° north of east into its horizontal and vertical components. This physics problem covers vector decomposition using trigonometry, ideal for students in Grades 9-12.

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