Explanation:
You have already determined the components of the known forces so I won't repeat your work here. Since the resultant force [tex]\vec{\textbf{R}}[/tex] and F1 are completely along the x-axis, we can conclude that
[tex]F_{2y} = F_{3y} \Rightarrow F_{3y} = F_3\cos{\theta} = 250\:\text{lb}[/tex]
We can now solve for the magnitude of [tex]F_3:[/tex]
[tex]F_3 = \sqrt{F_{3x}^2 + F_{3y}^2} = \sqrt{(467)^2 + (250)^2}[/tex]
[tex]\:\:\:\:=529.7\:\text{lb}[/tex]
The angle [tex]\theta[/tex] is then
[tex]\tan{\theta} = \dfrac{F_{3y}}{F_{3x}} = \dfrac{250}{467}[/tex]
or
[tex]\theta = 49.2°[/tex]
