A chair is pulled by two horizontal forces. The first force is 122 N at an angle of [tex]$43.6^{\circ}$[/tex], and the second is 97.6 N at an angle of [tex]$49.9^{\circ}$[/tex].

What is the [tex]$x$[/tex]-component of the total force acting on the chair?

[tex]$\overrightarrow{F_{x}} = [?] \, \text{N}$[/tex]

To find the [tex]$x$[/tex]-component of the total force acting on the chair, we need to consider each force separately and use trigonometry:

1. Identify the Forces and Angles:
- The first force is 122 N at an angle of 43.6° from the horizontal.
- The second force is 97.6 N at an angle of 49.9° from the horizontal.

2. Calculate the [tex]$x$[/tex]-component for Each Force:
- For the first force, the [tex]$x$[/tex]-component can be found using the cosine of the angle:
[tex]\[
\text{Force 1}_x = 122 \, \text{N} \times \cos(43.6^\circ)
\][/tex]
This calculation results in approximately 88.35 N.

- For the second force, similarly, we use the cosine of the angle:
[tex]\[
\text{Force 2}_x = 97.6 \, \text{N} \times \cos(49.9^\circ)
\][/tex]
This calculation results in approximately 62.87 N.

3. Combine the [tex]$x$[/tex]-components:
- The total [tex]$x$[/tex]-component of the force is the sum of the [tex]$x$[/tex]-components from both forces:
[tex]\[
\overrightarrow{F_{x}} = 88.35 \, \text{N} + 62.87 \, \text{N}
\][/tex]
This results in a total [tex]$x$[/tex]-component of approximately 151.22 N.

Therefore, the [tex]$x$[/tex]-component of the total force acting on the chair is about 151.22 N.