$∴Σ F_x (→ +ve)=R_x=50 \cos 60+120 \\ ∴R_x=120+25=145N (→) \\ ∴Σ F_y (↑ +ve)=R_y=-100- 50 \sin 60 \\ ∴R_y= 143.301N(↓) \\ \text {Resultant force (R) } =\sqrt{R_x^2+R_y^2} \\ = \sqrt{(145)^2+(143.301)^2 }=203.863N \\ R = 203.863N \\ θ= \tan^{-1}(\dfrac {R_y}{R_x} )= \tan^{-1}(\dfrac {143.301}{145}) \\ =44.66°$

Location of resultant force:

Let the resultant located at a perpendicular distance ‘d’ from B.( to produce anti - clockwise moment)

Using Varignon’s theorem,

$∴ ΣM_B^R= M_B^F ( +ve) \\ ∴R ×d=50 \sin60 × 40+100 ×20-120 ×40 \sin 60 \\ ∴d= \dfrac {424.871}{203.863 }=-2.08cm, \text {Hence , R is at right of B. }$