Explanation :
$6_{1,2}$ = $\begin{array}{l}\frac{6x+6y}2\pm\sqrt{\left(\frac{6x-6y}2\right)^2+(Txy)^2}\\\end{array}$
=80+20$\begin{array}{l}\frac{80+20}2\pm\sqrt{\left(\frac{80-20}2\right)^2+40^2}\\\end{array}$
=$\begin{array}{l}50\pm\sqrt{50^2}\\\end{array}$
$6_{1,2}$=100,0
As per Tresca's , max shear stress theory ,
T = $\begin{array}{l}\frac{6_1-6_2}2=\frac{100-0}2=50\\\end{array}$
F.O.S.= $\begin{array}{l}\frac{max\;shear\;of\;material}{max\;shear\;stress}=\frac{material\;strength}{design\;load}\\\end{array}$
100/50 = 2 .