Vectors of force and their angles This is a homework problem which has me stumped.  I have just begun Calculus 3 and this is the first section introducing vectors.  .
The embedded picture states the problem and shows some of my hen's scratching in an effort to solve it.  After studying the example problem from the same sub-chapter, I understand that the two vectors given, when summed, is the vector that is acting in opposition to the vector pulling the weight down.  That is $\mathbf{F_1} + \mathbf{F_2} = \langle 0,50 \rangle$.  This leads to the following system of equations:
$$
\begin{array}{rcl}
-\left|\mathbf{F_1}\right|cos(\alpha) + \left|\mathbf{F_2}\right|cos(60^\circ) & = & 0 \\
\left|\mathbf{F_1}\right|sin(\alpha) + \left|\mathbf{F_2}\right|sin(60^\circ) & = & 50 \\
\end{array}
$$
Basically, I'm having troubles figuring out how to eliminate one of the two unknowns: $\alpha$ and $\left|\mathbf{F_2}\right|$.  Using the system of equations, I can tell you that $\left|\mathbf{F_2}\right|$ is as follows:
$$
\begin{array}{rcl}
-\left|\mathbf{F_1}\right|cos(\alpha) + \left|\mathbf{F_2}\right|cos(60^\circ) & = & 0 \\
-35cos(\alpha) + \left|\mathbf{F_2}\right|\frac{1}{2} & = & 0 \\
-70cos(\alpha) + \left|\mathbf{F_2}\right| & = & 0 \\
\left|\mathbf{F_2}\right| & = & 70cos(\alpha) \\
\end{array}
$$
Also, knowing that the interior angles of a triangle must sum $180^\circ$, it's also quite a simple matter to demonstrate that $\alpha$ is as follows (assigning $\beta$ to the unnamed angle):
$$
\begin{array}{rcl}
180 & = & 60 + \alpha + \beta \\
\alpha & = & 120 - \beta \\
\end{array}
$$
From the scratching notes you can see in the picture, you can see that I've tried to draw in the unseen vector $\langle 0,50 \rangle$ in the hopes of making a right triangle to get myself closer to the answer.  This didn't help me.  I had the inspiration that some of the trigonometric functions and identities I learned (too many years ago now) would be of use.  So I pulled out my pre-calc book to look up things like SSA and so forth.  However, none seem to be the answer.  For example, the SSA only works if you know 2 of the three sides and one angle.  Well, I've got to of the three required criteria.
I also refreshed myself on the law of sines.  However, this doesn't bring me closer to (what I think are) the solutions.  So, one again, I can tell you this much:
$$
\text{The law of sines} \\
\frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c} \\
\text{I don't know C, or c, but this shouldn't matter} \\
\Rightarrow \frac{sin(60)}{35} = \frac{sin(\alpha)}{\left|\mathbf{F_2}\right|}
$$
However, it's pretty obvious that, though I can calculate the left side, it doesn't bring me close enough to get either $\alpha$ or $\left| \mathbf{F_2} \right|$
So, if someone could kindly point me in the right direction, I'd be very grateful.  Thanks.

Using Lami's there
$\frac{F_2}{sin (90+\alpha)}=2F_1=\frac{50}{sin (120-\alpha)}$
$\frac{F_2}{cos \alpha}=2F_1=\frac{50}{sin (120-\alpha)}$
$\frac{F_2}{cos \alpha}=\frac{50}{sin 120cos\alpha-cos 120sin\alpha}$
$\frac{F_2}{cos \alpha}=\frac{50}{\frac{\sqrt3}{2}cos\alpha+\frac{1}{2}sin\alpha}$
$\frac{F_2}{cos \alpha}=\frac{100}{\sqrt3cos\alpha+sin\alpha}$
Substituting your answer above where you got $F_2=70 cos \alpha$,
$70=\frac{100}{\sqrt3cos\alpha+sin\alpha}$
$\sqrt3cos\alpha+sin\alpha=\frac{10}{7}$
By using double angle,
R=2, $\theta=60$
$\sqrt3cos\alpha+sin\alpha=2sin(\alpha+60)$
$2sin(\alpha+60)=\frac{10}{7}$
$sin(\alpha+60)=\frac{5}{7}$
$\alpha+60=45.584$
$\alpha=-14.4
Since $60\le \alpha \le 120, \alpha=-14.4$
$F_2=67.79$.
I am getting the angle to be -ve, meaning sin is -ve and cos is +ve in the 4th quadrant. Angles in the 3rd quadrant too keep the sin -ve exceeds 120. It sounds a bit improbable to have a negative angle in real life, but according to the question, it makes sense.  
Good luck. 
