Problem 360 | Equilibrium of Non-Concurrent Force System

From the Force Polygon
$\dfrac{R_A}{\sin 45^\circ} = \dfrac{20 + T}{\sin 105^\circ}$

$R_A = 0.732(20 + T)$

$R_A = 14.641 + 0.732T$
 

From the Free Body Diagram
$\Sigma M_B = 0$

$4(R_A \cos 30^\circ) = 3(20) + 1(T)$

$4(14.641 + 0.732T)\cos 30^\circ = 60 + T$

$50.7179 + 2.5357T = 60 + T$

$1.5357T = 9.2821$

$T = 6.04 \, \text{ kN}$           answer