From the FBD of the whole system
ΣMD=0
2Ax=3(60)+6(30)
Ax=180 kN
ΣMA=0
2Dx=3(60)+6(30)
Dx=180 kN
From the FBD of member ABC
ΣFx=0
35FCF=Ax
35FCF=180
FCF=300 kN
ΣMB=0
3Ay=1.5(45FCF)
3Ay=1.2(300)
Ay=120 kN
From FBD of the whole system
ΣFV=0
Dy=Ay+60+30
Dy=120+60+30
Dy=210 kN
Summary
Ax=180 kN leftward
Ay=120 kN downward
Dx=180 kN rightward
Dy=210 kN upward
FCF=300 kN tension