Vector Tension Using Calculus 
A clothesline is tied between two poles, 8m apart. The line is quite
  taut and has negligible sag. When a wet shirt with a mass of 0.8kg is
  hung at the middle of the line, the midpoint is pulled down 8cm. Find
  the tension in each half of the clothesline.

I am trying to solve this problem similar to how this problem is solved:

The problem I am having is that I am getting the same angle for both tension formulas. I can show my work if needed.
Answer:

My work:

 A: There is one small mistake you have made, but you are on the right track. The $\theta$ value on the left of your picture (which I will call $\theta_l$) is not $-1.1457^\circ$. It's just $1.1457^\circ$. You can see this geometrically from the picture. Note that if you type 
$$\tan^{-1}{\frac{-8}{400}}$$
into your calculator it will give you a value in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$, which is why it gave you $-1.1457^\circ$ However, the tip of $\vec{T_1}$ is in the third quadrant so that vector is actually at an angle of $-1.1457^\circ + 180^\circ = 178.8543^\circ$ measured from the positive $x$ axis. So $\theta_l = 180^\circ - 178.8543^\circ = 1.1457^\circ$. Most of that is not necessary to go through though. You can just see from the symmetry of the situation that the angle on the left must equal the angle on the right.
Aside from this, your answer is so far correct. All you are trying to do is solve the system of equations at the bottom of the page. You have found from the equation
$$
-| \vec{T_1}|\cos{1.1457^\circ} + |\vec{T_2}|\cos{1.1457^\circ} = 0
$$
that $|T_1| = |T_2|$. You can now substitute $|\vec{T_1}|$ for $|\vec{T_2}|$ in the second equation, which would look like
$$
|\vec{T_1}|\sin{1.1457^\circ} + |\vec{T_1}|\sin{1.1457^\circ} = 0.8g
$$
You should be able to solve this and proceed. Note that I've made another small change to the right hand side of this last equation. You forgot to include multiplication by $g$, the acceleration due to gravity.
EDIT: The reason why they didn't bother multiplying by $g$ in the book is because for that problem the units were in pounds. A pound is a unit of force, whereas a kilogram is a unit of mass. You need to multiply the mass by $g$ to find the corresponding force in Newtons.
