## Engineering Mechanics: Statics & Dynamics (14th Edition)

$\theta=70.1^{\circ}$ $T=7.66kN$
First we will sum the forces horizontally $(\rightarrow +) \sum \overrightarrow{F}_{x} =0$ $T\cos\phi−4/5(8)=0$ Now we will sum the forces vertically $(\uparrow +) \sum \overrightarrow{F}_{y} =0$ $9-T\sin\phi-3/5(8)=0$ Solving the system of two equations, we obtain : $T=7.66kN$ $\phi=33.27^{\circ}$ $\phi$ is the angle that member C makes with the x-axis. We were asked for the angle between B and C, so we will need to add $\phi$ to the angle that member B makes with the x-axi. $\theta= \phi + \arctan(3/4)= 33.27^{\circ} + \arctan(3/4)=70.1^{\circ}$