Electrostatics 1st year engineering

[Dell]

at each corner of a rhobus , made of rods, each with a length of B, there are electric charges of "Q" and "q" so that each q charge is opposite a q charge and each Q charge opposite a Q charge. what does the ratio of Q/q need to be so that the angle (alpha) is

a)90 degrees
b)60 degrees
c)45 degrees
(take the tension of the rods into account)

i started up by taking one of the corners and analyzing the forces on it, since the shape is static, Ftotal=0.
i took the top Q charge and said, what are the forces acting on it,

the Coulumb force F(Q-Q) which pushes it upwards
the Coulumb force F(Q-q) which (depends on the charge)
the tension os the rods T

my problem is that i do not know which of the charges are negative(if any) or positive, so i cannot decide on the directions of the forces, the only one i do know is F(Q-Q) since Q must repell Q,

diagram below


diagram below
http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5311131135994735922

 

[Dell]

what i did so far is:

since i don't know the sign of q, the coulomb force can either repell or attract Q

F(QQ)=2Tcos(α/2) ± 2F(Qq)cos(α/2)

q will always repell q so T is the force which balance that
2Tsin(α/2) = F(qq)


F(QQ)=F(qq)cot(α/2) ± 2F(Qq)cos(α/2)

K$$\frac{QQ}{(2b*cos(alpha/2))^2}$$ = K$$\frac{qq}{(2b*sin(alpha/2))^2}$$*cot(alpha/2) ± 2K$$\frac{Qq}{b^2}$$cos(alpha/2)

(Q/q) * $$\frac{Q}{cos^2(alpha/2)}$$ = $$\frac{q}{sin^2(alpha/2)}$$*cot(alpha/2) ± 8Qcos(alpha/2)

1st of all, is this alright?
2nd of all, how do i find Q/q from this?? how do i get the q's and Q's out of this equation?