Equilateral triangle mechanics

1. The problem statement, all variables and given/known data

Three particles with masses m1=2.5kg, m2=1.2kg and unknown mass m3 are placed at the corners of an equilateral triangle of side 30.0cm. If the resultant force in x-direction acting on m1 is zero, find the net force acting on m1.





2. Relevant equations

F=(Gm1m2)/R^2

Fx=Fcosθ

Fy=Fsinθ

F=sqrt/(Fx)^2 +(Fy)^2



3. The attempt at a solution

Since Fx on m1=0

Fm1 cos 60 = 0

Fm1 = 0

The Fy on m1 = 0

Fy = Fm1 sin 60

Fy = 0 sin 60

Fy = 0

So does it mean the net force acting on m1 is

F=sqrt/(0)^2 +(0)^2

F=0N ?