Question

three forces of 10N,30N and 50N act at a point at 120° to each other.Find their resultant.

Answer 1

Assuming the 10N force is along the positive x-axis.

Then the three forces in (magnitude, angle °) are
P(10,0), Q(30,120), R(50,240)
or in component form:
P<10,0>
Q<30cos(120),30sin(120)>=<-15, 25.9808>
R<50cos(240),50cos(240)>=<-25, -43.301>
The resultant is therefore the sum of the three vectors in component form:
Resultant=<Fx,Fy>=<-30,-17.32>
The magnitude can be found using
|F|=√(Fx²+Fy²)
and direction using
atan2(Fy/Fx), or
atan(Fy/Fx) [+180 if Fx<0]

Answer 2

Fr = 10N.[0o] + 30N.[120o] + 50N.[240o]

X = 10 + 30*Cos120 + 50*Cos240 = -30 N.
Y = 30*sin120 + 50*sin240 = -17.3 N.

Tan A = Y/X = -17.3/-30 = 0.57735.
A=30o S. of W. = 210o CCW from +x-axis.

Fr = -30/Cos210 = 34.64 N. @ 210o CCW.