Answer
$T=7.20kN$
$F=5.40kN$
Work Step by Step
Given that $\theta=90^{\circ}$, we can solve for the angle that C makes with the x-axis, $\phi$.
$\phi=90^{\circ}-\arctan(3/4)=53.13^{\circ}$
First we will sum the forces horizontally
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$T\cos53.13^{\circ}−4/5(F)=0$
Now we will sum the forces vertically
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$9-T\sin53.13^{\circ}-3/5(F)=0$
Solving the system of two equations, we obtain :
$T=7.20kN$
$F=5.40kN$
