How can one conclude upward force is related to tension in this question? In preparation for the SAT Physics Subject Test, I have been doing extra practice problems. This one states:

A garment bag hangs from a clothesline. The tension in the clothesline is $10 \,\text{N}$ on the right side of the garment bag and $10 \,\text{N}$ on the left side of the garment bag. What is the mass of the garment bag?

The book from which this problem comes has a tendency to state questions unclearly, and some of its solutions have even had typos (which is really frustrating when using the book as a practice resource). 
I interpreted the problem as something like this, where $T_1$ and $T_2$ both equal $10 \,\text{N}$.

Then $$F_\text{w} = mg = T_1 \sin\theta_1 + T_2 \sin\theta_2 $$ so
$$m = \frac{T_1 \sin\theta_1 + T_2 \sin\theta_2}{g} = \frac{10 (\sin\theta_1 + \sin\theta_2 )}{g} $$
Since the SAT uses $g = 10 \,\text{m/s}^2$, this simplifies to $$m = \sin\theta_1 + \sin\theta_2$$
Also, if two strings have the same tension, then each string makes the same angle with the horizontal. Thus $\theta_1 = \theta_2,$ and it follows that 
$$m = 2\sin\theta_1 .$$
Since the angles were not given, this question cannot be solved, assuming I had interpreted the question correctly. 

For reference, this is the solution the book provided (which I can't make any sense of):
Total upward force on the garment bag is equal to the tension in the clothesline. Therefore, the magnitude of $T_{\text{total}}$ equals the garment bag's weight, $mg.$
$$T_{\text{total}} = F_\text{w}$$
$$F_\text{w} = mg$$
$$\implies m = \frac{T_{\text{total}}}{g}$$
$$\implies m = \frac{10 \,\text{N}}{10 \,\text{m/s}^2} = \boxed{1 \,\text{kg}}$$
What confuses me about this solution is how "total upward force on the garment bag is equal to the tension in the clothesline." I don't think the people who wrote this question would be so careless as to omit necessary information. I think if they didn't provide angles, then the question wasn't meant to be interpreted in the above manner. If anybody knows how the problem should be set up, please provide an explanation/diagram :)
Sorry for the lengthy post and thanks for reading.
 A: If both the ropes are of identical material, then the fact that they have equal tension in them means, by symmetry, that they make equal angles with horizontal, call it $\theta$. Then the only equation you have is $mg=2Tsin\theta$. So unless $\theta$ is given you cannot solve the problem. If you make, say, $\theta$ small so that the hanging ropes are close to being horizontal, then a large tension $T$ will be required in the ropes so that it has appropriate component along vertical direction to balance the hanging weight.
A: 
$$T_1/\sin \psi=T_2/\sin\psi=W_t/\sin\theta$$
$$10/\sin\psi=10/\sin\psi=W_t/\sin\theta$$
Therefore
$$10/\sin\psi=W_t/\sin\theta$$
$$W_t=F 
$$
$$\theta=360-2\psi$$
$$W_t= \frac{10\sin\theta}{\sin\psi}\tag{using lami's theorem}$$
$$ =\frac{\sin (360-2\psi)*10}{\sin\psi}$$
we know that $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$
so
$$\sin (360-\psi-\psi)=-sin \psi\cos \psi-\cos\psi\sin \psi$$
$$ =-2\sin \psi\cos \psi$$
Substituting 
$$W_t=\frac{-2\sin \psi\cos \psi*10}{\sin \psi}$$
$$=-20\cos \psi$$
$$
180>\psi>90$$
Therefore wt between 20N to 0N
So 10N is justifiable
