Engineering Mechanics: Statics & Dynamics (14th Edition)

$T=7.20kN$ $F=5.40kN$
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 : $T=7.20kN$ $F=5.40kN$