Physics\Trig Question I'm having trouble figuring out how I'm supposed to find the force vectors for this problem. I know I'm supposed to use trig somehow but I can't wrap my head around this since I have no sides just angles.

 A: Resolve the forces horizontally and vertically, which gives you a pair of simultaneous equations.
Vertically: 
$$F_1 \sin 30 + F_2 \sin 35 = 300$$
Horizontally: 
$$F_1 \cos 30  =  F_2 \cos 35 $$
Rearrange the second equation, to get $F_1$ in terms of $F_2$, and then substitute into the first equation
A: Hint:
As there is static equilibrium, the sum of the three forces ($F_1$, $F_2$ and the weight) is zero.
Obviously, if you consider the projections of the three forces on some axis, the sum is still zero. The trick is to choose the axis in a clever way to get some simplification.
For instance, if you project on an axis that is perpendicular to $F_2$, the $F_2$ contribution vanishes and you get an equation like
$$F_1\cos\alpha=W\cos\beta$$ where $\alpha$ and $\beta$ are angles that you need to determine.

A: I did what Henry L suggested and used my given knowledge of F1(sin30)+F2(sin35)=300 and F1(cos30)-F2(cos35)=0 to put one set of the system of equations in terms of just F2. The resulting equation was:
F2(cos(35)/cos(30))sin30+F2sin(35)=300 From that, I factored out F2 and got
F2((cos(35)/cos(30))sin30+sin(35))=300 I moved everything inside of the parenthesis into the other side
F2=300/((cos(35)/cos(30))sin30+sin(35))
The result was F2=286.67
From there I can apply the same logic to get F1. Once that is done I will have my sides and obtaining the vectors will be easy.
A: You can understand through pictures,

Using geometry,

Resolving vectors

I know it's not perfect but In the above picture the vertical vector represents both the resolutions of $F_1$ and $F_2$
Now you could just equate the forces

Therefore,
$$F_1\cos(30)=F_2\cos(35)$$
$$F_1\sin(30)+F_2\sin(35)=300$$
