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+//tension in the cable and reaction at a point
+//refer fig. 3.21 (a),(b)&(c)
+//Taking moment about A we get
+T=((25*12*cosd(30))+(10*6*cosd(30)))/(12*sind(15)) //kN
+//applying equilibrium conditions
+HA=T*cosd(15) //kN
+VA=10+25+T*sind(15) //kN
+RA=sqrt(HA^2+VA^2) //kN
+alpha=atand(VA/HA) //degree
+printf("tension T in cable BC is T=%.2f kN.\nRA=%.3f kN\nalpha=%.2f degree",T,RA,alpha)