F.B.D :-

Applying COE,

$ ∴Σ F_y (↑ +ve)=0 \\ ∴ -80+ V_B-100-60+ R_F \cos30=0 \\ ∴ V_B+ R_F \cos30=240…………(I) \\ ∴ ΣM_A \cong 0 ( +ve ) \\ ∴ -80 ×2+ V_B ×3-100 \sin45 ×5-50-60×7+ R_F \cos30×9=0 \\ ∴3V_B+ R_F \cos30×9=983.553…………….(II) $

Solving (I) and (II) simultaneously,

$∴ V_B=196.07 kN ,R_F=50.72kN \\ ∴Σ F_x (→ +ve)=0 \\ ∴H_B-100 \cos45- R_F \sin30=0 \\ ∴H_B=100 \cos45+50.72 \sin30 \\ ∴H_B=96.07 kN \\ R_B= \sqrt{H_B^2+V_B^2} =218.342 kN \\ θ= \tan^{-1}(\dfrac {V_B}{H_B })= \tan^{-1}(\dfrac {196.07}{96.07})= 63.90° $