To find the horizontal component $CA_x$ of the force $CA$, we use trigonometry. From the diagram, it looks like the force $CA$ has a direction along a 3-4-5 triangle, indicating that the angle between $CA$ and the horizontal axis is determined by the ratio of the opposite and adjacent sides (i.e., 3 and 4).

The horizontal component $CA_x$ can be calculated as follows:

$CA_x = CA \times \frac{4}{5}$

Given that $CA = 436.4 \, \text{N}$, we substitute:

$CA_x = 436.4 \times \frac{4}{5} = 436.4 \times 0.8$

Now, let's calculate that.The horizontal component $CA_x$ is $349.12 \, \text{N}$.

Do you have any further questions or would you like more details? Here are some related questions to deepen your understanding:

1. How would you calculate the vertical component of $CA$?
2. What is the significance of using a 3-4-5 triangle in this problem?
3. How do angles affect the calculation of force components?
4. What method would you use to find the magnitude of the resultant force if both components are known?
5. Can this same approach be applied to force $CB$, and what would be different?

Tip: Always ensure that the force vector components add up correctly using vector addition when resolving forces.