Mechanics Question, equilibrium

Hi, I would really appreciate any help on this question:
A particle P is attached to two points A and B by two light strings. The particle hangs in equilibrium. The tension in the string PA is 10N.
a.) Find the tension in the string PB
b.) the weight of the particle

Method would really help :)

Break down PA into vertical and horizontal forces (I'm assuming here that A makes an angle of 50 with the vertical.

PA Horizontal: 10cos(40)
PA Vertical: 10sin(40)

Then break the force exerted by PB (let's call that force ' T ') into vertical and horizontal components:

PB Horizontal: Tcos70
PB Vertical: Tsin70

Because the particle is in equilibrium the horizontal forces must cancel out:

PA Horizontal = PB Horizontal
10cos(40) = Tcos70
T = 10cos(40) / cos 70
T = 22.398 N

Note that we can't use the vertical forces here because there is a third force acting on the particle in a vertical plane - gravity. We can, however, now use this to find the particle's weight.

PA Vertical + PB Vertical = mg
10sin(40) + Tsin70 = 9.8m
m = [ 10sin(40) + 22.398sin70 ] / 9.8
m = 2.8kg

If my assumption for which angle is where is wrong just replace the angles in the appropriate places.

By symmetry PB = 10N (the horizontal components of PA and PB must be the same - this assumes that the strings are of equal length.)

You can't find the weight unless you are given the angles of the strings to the horizontal.