The catenary shape equation is $$ y(x) = y_C + a \left( \cosh \left( \frac{x-x_C}{a} \right) -1 \right) $$ where $(x_C,y_C)$ is the coordinate of the lowest point and $a$ is the so called catenary constant. At the supports the forces on the horizontal direction are $H$ and the unit weight of the cable is $w = \rho g A$ making the catenary constant $$a = \frac{H}{w} $$
Use the above to find the end-points of each segment. For a segment spanning between $x_1$ and $x_2$ in the horizontal direction with length $\ell$ and angle orientation $\theta$ the following is true
$$ \begin{align}
x_2 = x_1 + \ell \cos \theta \\
y_2 = y_1 + \ell \sin \theta
\end{align} $$
Now for the balance of forces. The weight of each segment is $W = w \ell$ and the force equations
$$ \begin{align}
(T + \Delta T) \cos \left( \theta + \Delta \theta\right) - T \cos \left( \theta \right) & = 0 \\
(T + \Delta T) \sin \left( \theta + \Delta \theta\right) - T \sin \left( \theta \right) - w \ell & = 0
\end{align} $$
where $T$ and $\theta$ is the tension and angle on the left side of the segment and $T+\Delta T$ and $\theta + \Delta \theta$ the tension and angle on the right side.
This means that each segment recursively changes the tension and angle by
$$ \begin{align}
\Delta T & = \sqrt{T^2+w^2 \ell^2+2 T w \ell \sin \theta}-T \\
\Delta \theta & = \tan^{-1} \left( \frac{\cos \theta}{\sin\theta + \frac{T}{w \ell} } \right)
\end{align} $$
You can check the results against the analytical form of the tension and angle
$$ \begin{align}
T &= H \cosh\left( \frac{x-x_C}{a} \right) \\
\theta &= \sinh\left( \frac{x-x_C}{a} \right)
\end{align} $$
