$ α= tan^{-1} (\dfrac45)= tan^{-1} (0.8)=38.66° \\ Now, \\ ∴Σ F_x (→ +ve)=R_x= 6\cos 38.66-20= -15.315N \\ ∴R_x=15.315N (←) \\ ∴Σ F_y (↑ +ve)=R_y=12 + 6 \sin 38.66= 15.748N(↑) \\ ∴\text { Resultant force (R)} = \sqrt{R_x^2+R_y^2 }= \sqrt{(-15.315)^2+(15.748)^2 }=21.367N \\ θ= \tan^{-1}(\dfrac {R_y}{R_x }) = \tan^{-1} (\dfrac {15.748}{-15.315})=45.79° $

Location of resultant force:

Let the resultant force R be located at a perpendicular distance ‘d’ from 0.

Using Varignon’s theorem,

$∴ ΣM_o^F= M_o^R ( +ve) \\ ∴20×2+12×3+15-20+35=R ×d \\ ∴d= \dfrac {20×2+12×3+15-20+35}{21.967 }= \dfrac {106}{21.967}\\ =4.825$

To find x- intercept :-

$ ∴ \sin⁡ 45.79°= \dfrac dx \\ ∴x= \dfrac d{\sin 45.79°} = \dfrac { 4.825} {\sin 45.79° }$

$= 6.73$ m to the right of O