Physics- Need today!!!?

Question

An object in equilibrium has three forces exerted on it. A 30 N force acts at 90° from the x-axis and a 44 N force acts at 60° from the x-axis. What are the magnitude and direction of the third force?

What is the third force(N)??
What is the degrees??° (counterclockwise from the +x direction)

Answer 1

So there is a third force!

The sum of all forces at equilibrium is equal to ...0 !
So
F1x+F2x +F3x=0
0 + Fcos(60) +F3x=0
44 cos(60) + Fx3=0
Fx3=44 cos(60)=22N

F1y+F2y +F3y=0
F1 + Fsin(60) + Fy3=0
30 +44sin(60) + Fy3=0
Fy3=-30 -44sin(60)= - 68.1N

F3=sqrt(Fx3^2 + Fy3^2)=71.6N
A=arcTan(Fy3/Fx3)=72.1

or A=180+ 72.1=252.1 from the positive x axis counter clockwise.