Let's use the point where the three strings meet as an object for a FBD.

There are three tensions, T1 acting up and to the right, T2 acting up and to the left, and T3 acting downward.

Okay, Newtons's 2nd.

Since there only two tensions acting in the horizontal direction, and that the point is stationary then we know

\(\displaystyle T_1 ~ cos( \theta _1 ) - T_2 ~ cos( \theta _2 ) = 0\)

In the vertical direction we get

\(\displaystyle T_1 ~ sin( \theta _1 ) + T_2 ~ sin( \theta _2 ) - T_3 = 0\)

Now, we know that \(\displaystyle T_1 = (1.50 ~ \text{kg} )g\) (why?) and likewise \(\displaystyle T_2 = (2.00 ~ \text{kg} )g\) and \(\displaystyle T_3 = (2.5 ~ \text{kg} )g\).

(Make sure you know why this is.)

So, you've got two equations (the Newton's Law equations) in two unknowns. Can you take it from here?

-Dan