Net force of two forces in a ring

[ShizukaSm]

Homework Statement

Consider the following ring:

Considering that $\theta = 30°$, determine F1 and F2 in order for the net force to be oriented downards with magnitude 10^3 N and without any horizontal component.

The Attempt at a Solution

Alright, so, the problem is I find two conflicting solutions when solving this simple problem.

Solution 1:
$$\\<br /> \sum F_x = 0 \rightarrow F_1*Sin(20)-F_2*Sin(30)=0 \\<br /> \sum F_y = 0 \rightarrow F_1*Cos(20)+F_2*Cos(30)=1000<br /> \\ \rightarrow F_2 ({\frac{Sin(30)}{Sin(20)} + Cos(30)}) = 1000<br /> \\ \rightarrow F_2 = 429.56N | F_1 = 627.91N$$
PS: How do I fix my parenthesis in line 3 of my latex equation? It's lacking proportion.

Solution 2(Book solution):

So... In short, why aren't those two solutions agreeing?

 

[haruspex]

$$\\<br /> \sum F_x = 0 \rightarrow F_1*Sin(20)-F_2*Sin(30)=0 \\<br /> \sum F_y = 0 \rightarrow F_1*Cos(20)+F_2*Cos(30)=1000<br /> \\ \rightarrow F_2 ({\frac{Sin(30)}{Cos(20)} + Cos(30)}) = 1000$$

Check that last step.

PS: How do I fix my parenthesis in line 3 of my latex equation? It's lacking proportion.

Try using \left( etc. instead of just (.

 

[ShizukaSm]

Check that last step.

Try using \left( etc. instead of just (.

Ahh, got it, can't believe I made such a stupid mistake. I checked and re-checked many times but couldn't spot it myself.

If anyone's wondering, it was supposed to be $\frac{F_1*(Sin(30)*Cos(20)}{Cos(20)}+...$