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+//Finding forces in members
+//Refer fig. 4.15 (a)
+RA=7*20/2 //kN
+RB=RA //kN (symmetry)
+//CE is perpendicular on AB
+CE=5.196 //m
+DE=3 //m
+theta=atand(5.196/3) //degree
+//The fact that 20 kN loads are equidistant can be used to find out horizontal distances of loads from A
+//Consider equilibrium of left hand side portion of section (1)-(1)
+//taking moment about A
+F2=(20*2.25+20*4.5+20*6.75)/(6*sind(60)) //kN (Tension)
+F1=(70-20-20-20+51.96*sind(60))/sind(30) //kN (compression)
+F3=-51.96*cosd(60)+110*cosd(30) //kN (Tension)
+printf("The required forces are:-\nF1=%.2d kN (Compression)\nF2=%.2d kN (Tension)\nF3=%.2d kN (Tension)",F1,F2,F3)