If the system is in equilibrium, then, it is necessary that the horizontal component of F1 equals F3. Then, you have:
[tex]\begin{gathered} F_1\cos \theta=F_3 \\ F_1\cos \theta=57 \end{gathered}[/tex]
In order to find the angle, consider the vertical components of the forces:
[tex]\begin{gathered} 20+F_1\sin \theta=60 \\ F_1\sin \theta=40 \end{gathered}[/tex]
Next, divide first and second equation to get tangent of the angle:
[tex]\begin{gathered} \frac{F_1\sin \theta}{F_1\cos \theta}=\frac{40}{57} \\ \tan \theta=\frac{40}{57} \\ \theta\tan ^{-1}(\frac{40}{57})\approx35 \end{gathered}[/tex]
Hence, the angle must be approximately 35 degrees.
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