Answer:
[tex]28.0\; {\rm N}[/tex] to the right.
Explanation:
Since the sign is not moving, the net force on this sign should be [tex]0\; {\rm N}[/tex]. For that, the horizontal component ([tex]x[/tex]-component) of external forces on this sign should be [tex]0\; {\rm N}[/tex].
Sources of external forces on this sign include tension from the wires, as well as gravitational pull (weight) from the earth. The gravitational pull from the earth is entirely vertical ([tex]y[/tex]-component,) with a magnitude of [tex]0\; {\rm N}[/tex] in the horizontal direction. Thus, the only external forces on this sign in the [tex]x[/tex]-component would be from the two wires.
The question states that the [tex]x[/tex]-component of the force from the first wire is [tex]28.0\; {\rm N}[/tex] to the left. Thus, for the net force in the [tex]x[/tex]-direction to be [tex]0\; {\rm N}[/tex], the force from the other wire in the [tex]x\![/tex]-component needs to be [tex]28.0\; {\rm N}\![/tex] to the right (same magnitude but opposite direction.)
