Answer:
Explanation:
Given
Two forces [tex]F_1[/tex] and [tex]F_2[/tex] at an angle of [tex]\theta _1[/tex] [tex]\theta _2[/tex]
[tex]F_1=56\ N[/tex]
[tex]F_2=27\ N[/tex]
[tex]\theta _1=53^{\circ}[/tex]
[tex]\theta _2=156^{\circ}[/tex]
As resultant force is zero therefore horizontal component as well as vertical component of force is zero
[tex]\sum F_x=F_1\cos \theta _1+F_2\cos \theta _2+F_3\cos \theta _3[/tex]
[tex]\sum F_x=56\cos 53+27\cos 156+F_3\cos \theta [/tex]
[tex]F_3\cos \theta_3=-9.035\ N----1[/tex]
[tex]\sum F_y=F_1\sin \theta _1+F_2\sin \theta _2+F_3\sin \theta _3[/tex]
[tex]F_3\sin \theta _3=55.704\ N----2[/tex]
squaring and adding 1 and 2
[tex]F_3^2(\cos ^2\theta _3+\sin^2\theta _3)=86.583+3102.93[/tex]
[tex]F_3=56.47\ N[/tex]
Divide 1 and 2 to get [tex]\theta _3[/tex]
[tex]\frac{\sin \theta_3 }{\cos \theta_3 }=\frac{-55.704}{9.305}[/tex]
[tex]\tan \theta_3=-6.16[/tex]
[tex]\theta _3=180-80.78[/tex]
[tex]\theta _3=99.22^{\circ}[/tex]