Question: Breaking strength rope question

Although a very general question... does the rope break or not?

A 100 kg weight is hung from the center of a 1/4" polypropelene rope making an angle of 10 degrees at both ends with the horizontal.  According to some characteristics, the breaking strength of the rope is 1125 lbs or 5kN.  The safe load (safety factor of 12) drops the weight to  93.8 lbs or 0.417kN. Find the tension in the rope?  Is it strong enough?

Fairly easy to solve in Maple - just two equations two unknowns making t1 tension on one side and t2 tension on the other side

eq1 := t1*cos(80*Pi*(1/180))+t2*cos(80*Pi*(1/180)) = 100*9.8 # ΣFy=mg

eq2 := t1*sin(80*Pi*(1/180))-t2*sin(80*Pi*(1/180)) = 0 #ΣFx=0 hence basically t1=t2

solve([eq1,eq2])

                 {t1 = 2821.797537, t2 = 2821.797537}

Therefore the tension on the rope t1=t2 is 2821 N  (looks like way beyond the safety factor) so this means the rope will not break, correct?  or is it that the total tension becomes t1+t2=5642 N putting us over the 5kN and the rope breaks?