How do I calculate the tension in cable 1 without using the tension in cable 2?

[ijd5000]

A chandelier with mass m is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension T1 and makes an angle of θ1 with the ceiling. Cable 2 has tension T2 and makes an angle of θ2 with the ceiling.

Find an expression for T1, the tension in cable 1, that does not depend on T2.
Express your answer in terms of some or all of the variables m, θ1, and θ2, as well as the magnitude of the acceleration due to gravity g.


i have for my Forces:

F(x) T2 cos ( θ2)-T1cos(θ1)=0
F(y) T2 sin(θ2)+T1 sin(θ1)-gm=0

not sure how to describe T1 without using T2

 

[arildno]

These are two equations in two unknowns, T_1 and T_2.
If you write them as
T_1cos(v_1)=T_2cos(v_2)
T_1sin(v_1)-mg=-T_2sin(v_2)
then dividing the one equation with the other yields:
\frac{T_{1}\sin\theta_{1}-mg}{T_{1}\cos\theta_{1}}=-\tan\theta_{2}

Hurray, T_2 has disappeared!
Now, rearrange the last equation, so as to solve for T_1

 

[ijd5000]

These are two equations in two unknowns, T_1 and T_2.
If you write them as
T_1cos(v_1)=T_2cos(v_2)
T_1sin(v_1)-mg=-T_2sin(v_2)
then dividing the one equation with the other yields:
\frac{T_{1}\sin\theta_{1}-mg}{T_{1}\cos\theta_{1}}=-\tan\theta_{2}

Hurray, T_2 has disappeared!
Now, rearrange the last equation, so as to solve for T_1

It's ok to combine equations from different directions? I thought you have to keep x and y separate until you add the vectors.