Answer:
Force will be 70 N so option (b) is correct option
Explanation:
We have given horizontal component of the force
[tex]F_x=60.6N[/tex]
Angle [tex]\Theta =30^{\circ}[/tex]
We have to find the net force
We know that horizontal force is given by [tex]F_X=Fcos\Theta[/tex]
So force F [tex]=\frac{60.6}{cos30}=\frac{60.6}{0.866}=69.97= 70 N[/tex]
So option (b) is correct option