The resultant of two equal forces is $141.4\mathrm{N}$  when they are mutually perpendicular. When they are inclined at an angle ${120}^0$ , then the resultant force will be 

Let each force be F then the resultant of these two forces for an angle ${90}^0$ is given as $141.4N$

$\therefore R=\sqrt{F^2+F^2+2FF\cos {90}^0}$

$141.4=\sqrt{2}F\Rightarrow 141.4=141.4F$

$\therefore F=100N$

Now for an angle $\theta ={120}^0,$ then resultant force will be $R_1=\sqrt{F^2+F^2+2FF\text{\hspace{0.17em}cos}{120}^0}$

$R_1=F(\because \cos {120}^0=\frac{-1}{2})$

$\therefore R_1=100N$