$\Sigma M_A = 0$
$24R_B + 3(20) + 3(\frac{1}{\sqrt{5}})(22.4) = 18(\frac{2}{\sqrt{5}})(22.4) + 18(30) + 12(20) + 6(10)$
$R_B = 46.27 \, \text{ kN}$ answer
$\Sigma M_B = 0$
$24A_V = 3(20) + 3(\frac{1}{\sqrt{5}})(22.4) + 6(\frac{2}{\sqrt{5}})(22.4) + 6(30) + 12(20) + 18(10)$
$A_V = 33.76 \, \text{ kN}$
$\Sigma F_H = 0$
$A_H = 20 + \frac{1}{\sqrt{5}}(22.4)$
$A_H = 30.02 \, \text{ kN}$
$R_A = \sqrt{{A_H}^2 + {A_V}^2}$
$R_A = \sqrt{30.02^2 + 33.76^2}$
$R_A = 45.18 \, \text{ kN}$
$\tan \theta_{Ax} = \dfrac{A_V}{A_H}$
$\tan \theta_{Ax} = \dfrac{33.76}{30.02}$
$\tan \theta_{Ax} = 48.36^\circ$
Thus, $R_A = 45.18 \, \text{ kN}$ up to the right at $48.36^\circ$ from horizontal. answer
