From right triangles ACD and ACB.
$\cos 30^\circ = \dfrac{AC}{6} = \dfrac{AC}{AB}$
$AB = 6 \, \text{ m}$
Notice also that triangle ABD is an equilateral triangle of sides 6 m.
$\Sigma M_A = 0$
$6(R_B \cos 30^\circ) = 3(10) + 6(10) + 9(10)$
$R_B = 20\sqrt{3} \, \text{ kN}$
$R_B = 34.64 \, \text{ kN}$ answer
$\Sigma F_H = 0$
$A_H + 4(10 \sin 30^\circ) = R_B \cos 30^\circ$
$A_H + 4(10 \sin 30^\circ) = 20\sqrt{3} \cos 30^\circ$
$A_H = 10 \, \text{ kN}$
$\Sigma F_V = 0$
$A_V + R_B \sin 30^\circ = 4(10 \cos 30^\circ)$
$A_V + 20\sqrt{3} \sin 30^\circ = 4(10 \cos 30^\circ)$
$A_V = 10\sqrt{3} \, \text{ kN}$
$A_V = 17.32 \, \text{ kN}$
$R_A = \sqrt{{A_H}^2 + {A_V}^2}$
$R_A = \sqrt{10^2 + (10\sqrt{3})^2}$
$R_A = 20 \, \text{ kN}$
$\tan \theta_{Ax} = \dfrac{A_V}{A_H}$
$\tan \theta_{Ax} = \dfrac{10\sqrt{3}}{10}$
$\theta_{Ax} = 60^\circ$
Thus, $R_A = 20 \, \text{ kN}$ up to the right at $60^\circ$ from horizontal. answer
