As Shown in FBD,

Taking two components of $R_A$ & taking support reaction at pt. B which inclined at angle $60^\circ$ with horizontal. Also converting the trapezoidal load into equivalent point loads we get FBD

Now , applying conditions of equilibrium

$\sum F_x=H_A- R_B \cos 60 = 0 ----- (1) \\ \sum F_Y =V_A- 120-180-80 + R_B \sin 60=0 ----(2) \\ \sum MFA \cong 0 [ \space\space\space\space\space +ve] \\ 120(5) +180(6)-R_B \sin 60 (10) +80(13)=0 \\ R_B=314.08 \space KN \space \space at \space \space \theta =60^\circ $

Putting in eqs. (1) and (2), we get

$H_A=157.04 KN , V_A =108 KN\\ So \space , R_A= 190.59 KN \space \space at \theta =34.52^\circ $

Support reaction at A is $190.59 KN$ at $\theta=34.52^\circ $ & at B is $R_B= 314.08 KN $ at $\theta=60^\circ$