#### Answer

$\theta=70.1^{\circ}$
$T=7.66kN$

#### Work Step by Step

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.
$\theta= \phi + \arctan(3/4)= 33.27^{\circ} + \arctan(3/4)=70.1^{\circ}$