Angles in a static equilibrium I have three masses $\left(F_\alpha, \, F_\beta , \, \text{and}  \,F_g \right)$ with 2 pulleys, and a wind variable which is in static equilibrium. I have already calculated the appropriate forces for the 3 masses by multiplying it with $9.81 \, \frac{\mathrm{m}}{\mathrm{s}^{2}}$ (gravity). 
$$
\begin{alignat}{7}
& F_{\text{wind}} && ~=~ & 60 \phantom{.0} & \, \mathrm{N}  \\[2px]
& F_{\alpha} && = & 313.9 & \, \mathrm{N} \\[2px]
& F_{\beta} && = & 619 \phantom{.0} & \, \mathrm{N} \\[2px]
& F_{g} && = & 882.9 & \, \mathrm{N} \\
\end{alignat}
$$
I'm required to find the angles for vector $F_\alpha$ and $F_\beta$ as shown in below equations (which is derived from the vector's individual components ($x$ and $y$):
$$
\begin{alignat}{7}
F_α \, \cos{\left( α \right)} & \, + \, F_β \, \cos {\left( β \right)} && + F_\text{wind} & ~=~ 0  \tag{1} \\[2px]
F_α \, \sin{\left(α\right)} & \, + \, F_β \, \sin {\left( β \right)} && - F_g & ~=~ 0  \tag{2}
\end{alignat}
$$
Replacing these with actual values: 
- 313.9cos α + 619cos β + 60 = 0  — (1)
 313.9sin α + 619sin β - 882.9 = 0  — (2)

How do I find the angle α & β from these two equations?
Edit 2:

I have re-organized the equation and square it as such:
cos²a = (619² cos²β + 60² + 2(619cosβ * 60)) / 313.9²
sin²a = (619² sin²β + 882.9² - 2(619sinβ * 882.9)) / 313.9²

You can eliminate the angle $\alpha$ from the equations with the trick the other answers give you *. But then you will end up with an equation of the form
$$ A \cos \beta + B \sin \beta + C = 0$$
To solve this do the following transformation
$$ \left. \begin{align}
  A & = R \cos \psi \\
  B & = R \sin \psi 
\end{align} \right\} 
\begin{aligned} 
  R & = \sqrt{A^2+B^2} \\
  \psi & = \arctan\left( \frac{B}{A} \right) 
\end{aligned}  $$
The equation is now $$ 
cos\beta\cos\psi + \sin\beta \sin\psi = \cos(\beta-\psi) = -\frac{C}{R} $$
which is solved for
$$ \begin{split} \beta & = \arccos\left( -\frac{C}{R} \right) + \psi \\
 & = \arccos\left( -\frac{C}{\sqrt{A^2+B^2}} \right) + \arctan\left( \frac{B}{A} \right)\end{split}$$
footnotes:


*

*make the equations of this form 
$$\begin{align} 
  \cos \alpha & = a \cos\beta+c_x \\
  \sin \alpha & = -a \sin \beta + c_y 
\end{align}$$

*square both sides and add them for
$$ 1 =  2 a c_x \cos\beta - 2 a c_y \sin\beta + c_x^2 + c_y^2 +a^2 $$
$$ \left(2 a c_x\right) \cos\beta + \left(- 2 a c_y\right) \sin\beta + \left(c_x^2 + c_y^2 +a^2-1\right) = 0 $$

*Match the $A$, $B$ and $C$ coefficients.

*Once $\beta$ is known, then divide the two equations above for $$ \tan \alpha = \frac{c_y - a \sin\beta}{c_x + a \cos\beta} $$


Edit 1
Here is the actual solution:
$$\left. \begin{align}
  -313.9 \cos(\alpha) + 619 \cos(\beta) + 60 & = 0 \\
   313.9 \sin(\alpha)  + 619 \sin(\beta) - 882.9 & = 0 
\end{align} \right\} \begin{aligned}
   313.9 \cos(\alpha) & = 619 \cos(\beta) + 60 \\
   313.9 \sin(\alpha)  & = - 619 \sin(\beta) + 882.9
\end{aligned} $$
Square and add the two equations (on each side) to get 
$$ \left. 98533.21 = 74280 \cos(\beta) - 1093030.2 \sin(\beta) + 1166273.41 \right\}\\ 74280 \cos(\beta) - 1093030.2 \sin(\beta) + 1067740.2 = 0 $$
$$ \begin{aligned} \beta & = \arccos\left( -\frac{C}{\sqrt{A^2+B^2}} \right) + \arctan\left( \frac{B}{A} \right) \\
A & = 74280\\
B & = -1093030.2 \\
C & = 1067740.2\\
\beta &= 1.41284652 = 80.9501426° \\ \end{aligned} $$
Finally, $\alpha$ can be solved with the 2nd equation:
$$ \sin(\alpha) = 2.81267919-1.97196559 \sin(\beta) $$
$$ \alpha = 1.04567064 = 59.9125144° $$
Now you can plug the values of $\alpha$ and $\beta$ into the two original equations to confirm it balances the forces.
