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Please refer the following images for the solution. Solution:Taking moment about A, sumM_(A)=0;{:[F_(B)xx32=(600 xx24)+(1200 xx16)+(800 xx8)],[+(300 xx32)],[:.F_(B)=1550lb]:}Taking resolution of vertical forces,SigmaF_(y)=0;{:[F_(A)+F_(B)=300+800+1200+600+300],[F_(A)=300+800+1200+600+300-1550],[:.F_(A)=1650lb]:}Using method of joints.At joint A,{:[AF_(AB)quad theta=tan^(-1)((6)/(8))],[F_(A)=1650bb]:}Taking resolution of vertical forces, +i sum Fy=0;{:[1650+F_(AB)xx sin theta-300=0],[:.F_(AB)=(300-1650)/(sin 36.87^(@))=-2250lb],[:.F_(AB)=2250lb" (compression) "]:}Taking resolution of horizontal forces, +-_(x)SigmaF_(x)=0{:[-F_(AB)xx cos theta+F_(AC)=0quad(F_(AB):}],[:.F_(AC)=F_(AB)xx cos 36.87^(@)],[:.F_(AC)=2250 xx cos 36.87^(@)],[F_(AC)=1800lb(" Tension) "]:}(FAB compression)At joint C,Taking resolution of vertical ferces, +iF_(y)=0quad FCB=0Taking resolution of horizantal farces,{:[I_(A)^(')SigmaF_(X)=0;F_(E)=F_(AC)=1800lb],[:.F_(CE)=1800lbquad" (Tension) "]:}At joint B.{:[soolb_("Fod since ")],[F_(BD)quadQ_(1)=36.87^(@)],[Q_(2)=tan^(-1)((2.3333)/(8))],[:.theta_(2)=16.26^(@)],[theta_(3)=tan^(-1)((6)/(8))],[:.theta_(3)=36.87^(@)],[:.SigmaF_(y)=0],[F_(AB)xx sin theta_(1)+F_(BD)*sin theta_(2)-F_(BE)xx sin theta_(3)-800=0],[(2250 xx sin 36.87)+F 60 xx sin 16.26-F 06 xx sin 36.87=800],[:.0.28F_(B)D-0.6F_(BE)=-550-0.28],[" Dividged by "0.28", "],[FBD-2.1428F_(BE)=-1964.28],[sumF_(x)=0"; "],[(2250 xx cos 36.87)+(F_(BD)xx c ... See the full answer