F.B.D:

Applying static conditions of equilibrium,

$∴Σ F_y (↑ +ve)=0 \\ -200-200+ R_B \cos10=0 \\ ∴ R_B \cos10=400 \\ ∴R_B= \dfrac {400}{\cos10 }=406.17 N (↑) \\ \cos10 = 50/x , ∴x=\dfrac {50}{\cos10}=50.77 mm \\ \tan10 = y/50 , ∴ y = 50 \tan10 = 8.816mm $

$∴ $ Distance of $R_A$ from $O=x+y=50.77+8.816=59.586 mm$

Now, $z= \sqrt{100^2-50^2 }$………{By Pythagoras Theorem}

$$ ∴z=86.60 mm$$

$∴$ Distance of $R_C$ from $O= 59.586+86.60=146.18 mm$ Perpendicular distance of $R_B $ from $O \rightarrow$ $$\text { Consider a triangle } ∆OAC : \ OC = \sqrt{OA^2+AC^2}= \sqrt{59.586^2+50^2 }=77.78 mm$$

$∴ \text { In right angled } ∆OBC , OB = \sqrt{77.78^2-50^2 }=59.586 mm \\ ∴Σ F_x (→ +ve)=0 \\ ∴ R_A- R_C- R_B \sin10=0 \\ ∴ R_A- R_C=406.17 \sin10=70.53…………………(I) \\ ∴ ΣM_O \cong 0 ( +ve ) \\ ∴ -R_A ×59.586+ R_B ×59.586-200×50-200×100+ R_C×146.18=0 \\ ∴-59.586 R_A+146.18 R_C =200×50+200×100=406.17 ×59.586 $

Solving equations (I) and (II) simultaneously, we get

$R_A=186.02 N ,R_C=115.488 N $