Help, static equilibrium solving for maximum angle of incline

[Runaway]

Homework Statement

A machine in an ice factory is capable of exerting 3.3 * 10^2 N of force to pull large blocks of ice up a slope. The blocks each weight 1.21 *10^4 N. Assuming there is no friction, what is the maximum angle that the slope can make with the horizontal if the machine is to be able to complete the task.
Fg= 1.21*10^4 N
Ft= 3.3 * 10^2 N
Fn is perpendicular to Ft

Homework Equations

(in case the symbols my teacher uses are not standard)
Fg= weight
Ft= tension force
Fn= normal force
F()x= the x component of said force ie Ftx
F()y= the y component of said force ie Fty

The Attempt at a Solution

These are the only equations I can come up with using the given data, but no matter how I substitute them, I have yet to find a solvable equation.
Ftx= Ft/sin($$\theta$$)
Fty=Ft/cos($$\theta$$)
Fnx=Fn/sin(90-$$\theta$$)
Fny=Fn/cos(90-$$\theta$$)
Ft= sqrt(Ftx² + Fty²)
Fn= sqrt(Fnx² + Fny²)
Fty+Fny=Fg
Ftx=Fnx
but no matter how I substitute them, I have yet to find a solvable equation.
ex.
1.21*10^4 N=(3.3 * 10^2 N)/cos($$\theta$$)+(sqrt(((3.3 * 10^2 N)/sin($$\theta$$)² + ((1.21*10^4 N)-(3.3 * 10^2 N)/cos($$\theta$$))²))/cos(90-$$\theta$$)

 

[The legend]

when the object is on an incline(without friction and with the machine pulling it) 2 forces act on it along the incline... i.e: the component of gravity and the pulling of the machine.

So equating those two you can get your answer..

PS:
There is no need of so many equations you've written up there..

 

[Runaway]

Thanks, I was completely over complicating the problem,
Am I looking at it correctly now,because I keep getting an answer of less than 2 degrees, which doesn't sound right?

Ft=cos(90-x)*Fg
-(cos^-1(Ft/Fg)-90)=x
-(cos^-1(3.3*10^2/1.21*10^4)-90)=1.563 degrees

 

[The legend]

yes looks right...

 

[Runaway]

thanks for your help :)

 

[The legend]

my pleasure..