Answer
$\theta=4.69^{\circ}$
$F_1=4.31kN$
Work Step by Step
First we will sum the forces horizontally
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$6\sin70^{\circ}+F_1\cos\theta−5\cos30^{\circ}−4/5(7)=0$
Now we will sum the forces vertically
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$6\cos70^{\circ}+5\sin30^{\circ}−F_1\sin\theta−3/5(7)=0$
Solving the system of two equations, we obtain :
$\theta=4.69^{\circ}$
$F_1=4.31kN$
