Chapter 3 - Equilibrium of a Particle - Section 3.3 - Coplanar Force Systems - Problems - Page 98: 3

$\theta=82.2^{\circ}$ $F=3.96kN$

First we will sum the forces horizontally $(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $ $F\sin\theta−4\cos60^{\circ}-8\cos30^{\circ}=0$ Now we will sum the forces vertically $(\uparrow +) \sum \overrightarrow{F}_{y} =0 $ $8\sin30^{\circ}-4\sin60^{\circ}−F\cos\theta=0$ Solving the system of two equations, we obtain : $\theta=82.2^{\circ}$ $F=3.96kN$