Mass on 2 ropes, looking for tension

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Homework Statement



A block of mass 3.3 kg is suspended by two ropes as shown in the picture below.
The angle that the left rope makes with the horizontal is q= 40 degrees.
The angle that the right rope makes with the horizontal is q= 30 degrees.

What is the tension in the left rope?
A. T1 = 16.5 N

B. T1= 22.4 N

C. T1= 29.8 N
What is the tension in the right rope?

A. T2 = 26.4 N

B. T2 = 30.2 N

C. T2 = 33.1 N

Homework Equations



F=ma

The Attempt at a Solution


I separated each of the T vectors into their x and y components and was planning on using the sine of the given angles in order to solve for the tension. I keep coming up with something like T=mg/sin30, but the answer that I get for this is nowhere near any of the options. I know I am probably missing something so simple, but I can't for the life of me figure out what it is. Please help!
 
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In the x direction I came up with T2x-T1x=0
In the y direction I came up with T2y+T1y=mg
I found mg=32.34
Not sure where to go from here though.
 
Ok, so T1x=T1*cos40? The problem with this is that I end up with 2 variables and I'm not sure how to solve for either one of them with the given info. It seems like the 32.34 should be worked in somewhere but being that it's the sum of T2y and T1y, I'm not sure how to apply it.
 
Thanks for the help! Still working out the details but it is good to know I'm on the right track.