The question asks for the horizontal component of Vector A. The information provided is:

• Vector A = 60 N, 245 degrees from the +X-axis.

To find the horizontal component of a vector given its magnitude and direction, we use the following formula:

$A_x = A \cdot \cos(\theta)$

Where:

• $A_x$ is the horizontal component of vector A.
• $A$ is the magnitude of vector A (60 N).
• $\theta$ is the angle of the vector measured counterclockwise from the +X-axis (245 degrees).

Step-by-step calculation:

1. First, recognize that the angle of 245 degrees places the vector in the third quadrant of the coordinate system, meaning the horizontal component will be negative.
2. Compute the horizontal component:

$A_x = 60 \cdot \cos(245^\circ)$

Let me calculate that for you.The horizontal component of Vector A is approximately $-25.4$, rounded to 3 significant figures.

Would you like further clarification or details on any part of this process?

Here are 5 related questions for further practice:

1. What is the vertical component of Vector A?
2. How would you find the resultant vector of Vector A and Vector B?
3. What is the magnitude and direction of Vector C?
4. How do you add vectors using the analytical method?
5. How do you determine the quadrant a vector is in based on its angle?

Tip: Remember that when the angle is measured from the positive X-axis, the cosine gives the horizontal component, and sine gives the vertical component!