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+// Problem no 10.3,Page No.252
+
+clc;clear;
+close;
+
+//taking moment at pt A we get
+R_D=(90*6+120*3)*9**-1 //Reaction at Pt D
+
+//Joint D
+
+//Applying the summation of vertical forces we get
+F_GD=100*(sin(60*%pi*180**-1))**-1
+
+//Applying the summation of horizontal forces we get
+F_DC=F_GD*cos(60*%pi*180**-1)
+
+//Joint G
+
+//Applying the summation of vertical forces we get
+F_GC=F_GD
+
+//Applying the summation of horizontal forces we get
+F_FG=F_GD*cos(60*%pi*180**-1)+F_GC*cos(60*%pi*180**-1)
+
+//joint C
+
+//Applying the summation of vertical forces we get
+F_FC=(115.5*sin(60*%pi*180**-1)-90)*(sin(60*%pi*180**-1))**-1
+
+//Applying the summation of horizontal forces we get
+F_CB=F_DC+F_GC*cos(60*%pi*180**-1)+F_FC*cos(60*%pi*180**-1)
+
+//joint F
+
+//Applying the summation of vertical forces we get
+F_FB=F_FC
+
+//Applying the summation of horizontal forces we get
+F_EF=F_FG+F_FC*cos(60*%pi*180**-1)+F_FB*cos(60*%pi*180**-1)
+
+//Joint B
+
+//Applying the summation of vertical forces we get
+F_EB=(120-F_FB*sin(60*%pi*180**-1))*(sin(60*%pi*180**-1))**-1
+
+//Applying the summation of horizontal forces we get
+F_BA=F_CB+F_FB*cos(60*%pi*180**-1)-F_EB*cos(60*%pi*180**-1)
+
+//Joint E
+
+//Applying the summation of vertical forces we get
+F_AE=F_EB
+
+//Result
+printf("Forces in Each members are as follows:F_GD %.1f kN (compression)",F_GD)
+printf("\n :F_DC %.2f",F_DC);printf(" KN(Tension)" )
+printf("\n :F_GC %.1f kN (Tension)",F_GC)
+printf("\n :F_FG %.1f kN (Compression)",F_FG)
+printf("\n :F_FC %.1f kN(compression)",F_FC)
+printf("\n :F_CB %.2f",F_CB);printf(" KN(Tension)")
+printf("\n :F_FB %.1f kN(compression)",F_FB)
+printf("\n :F_EF %.2f",F_EF);printf(" KN(compression)")
+printf("\n :F_EB %.2f",F_EB);printf(" KN(Tension)")
+printf("\n :F_BA %.2f",F_BA);printf(" KN(Tension)")
+printf("\n :F_AE %.2f",F_AE);printf(" KN(compression)")