Applying static conditions of equilibrium,

$∴Σ F_x (→ +ve)=0 \\ ∴ H_A-30 \cos 40=0 \\ ∴ H_A=30 \cos40=22.98 kN (→) \\ ∴ ΣM_A \cong 0 ( +ve) \\ ∴20 ×2 -25×4-30\sin40×8+30\cos40×1.5+R_B ×10=0 \\ ∴R_B ×10=40+100+154.269-34.472 \\ ∴R_B=(259.797)/10=25.978 kN (↑) \\ ∴Σ F_y (↑ +ve)=0 \\ ∴ V_A-20-25-30 \sin40+R_B=0 \\ ∴ V_A=20+25+19.284-25.978=38.306 kN (↑) \\ ∴ R_A= \sqrt{H_A^2+V_A^2 }= \sqrt{(22.98)^2+(38.306)^2}=44.67kN \\ θ= \tan^{-1}⁡(\dfrac {V_A}{H_A} )= \tan^{-1}⁡(\dfrac {38.306}{22.98})=59.04°$