Mass hanging in between two ropes - tensions of ropes

Hello, I would like some help with this problem -
Body shown is in equilibrium. Find tensions T1 and T2 in the below picture

http://i49.tinypic.com/1iz48.png

I know up to this point ..

Along X : T1 Cos(45*) = T2Cos(30*)
Along Y : T1Sin(45*) = T2Sin(30*)
Do not know what to do after.

Thank you.

In the x-direction, forces are acting in opposite directions, so letting left be negative and right be positive:

-T1 *cos(45) + T2 * cos(30) = 0
T1 * cos(45) = T2 * cos(30)

In the y-direction, both ropes pull up and the mass pulls down, so:

T1 * sin(45) + T2 * sin(30) - m*g = 0
T1 * sin(45) + T2 * sin(30) = (20 kg)(9.81)

Solve for T1 or T2 in the first equation and get:

T1 = [T2 * cos(30)]/cos(45)

Plug it into your second and get:

[T2 * cos(30)]/cos(45) * sin(45) + T2 * sin(30) = (20 kg)(9.81)
T2 * cos(30) * tan(45) + T2 * sin(30) = (20 kg)(9.81)
[cos(30)*tan(45) + sin(30)]T2 = (20)(9.81)
T2 = (20)(9.81)/[cos(30)*tan(45) + sin(30)]
T2 = 143.628 N.

T1 = [T2 * cos(30)]/cos(45)
T1 = 175.908 N

Looks like you fudged the substitution part.

There is a mistake with the equation along Y. Gravity pulls down on the block, so it should be T1sin(45)+T2sin(30)-mg = 0.