equilibrium

Yes.  At least, that's what I've assumed!

.

I can't be sure from the limited information in your question, but perhaps you are thinking of something like the following:

If so, then resolve forces horizontally and vertically:

Horizontal:

T1cos(15°) = T2cos(10°) 

Vertical:

T1sin(15°)+T2sin(10°) = 300

You now have two equations in two unknowns, so you can solve for T1 and T2.

You didn't give the units for the weight. If you meant the 300 to represent a mass, you will have to multiply it by g, the gravitational acceleration.

Edit:  I've just noticed there is a degree symbol (°) attached to T2 in my picture!  This is a misprint and should be ignored.

.

Hi Alan,

What does T stand for ? (Tension?)

Thanks Alan.