The problem asks to find the angles among $420^{\circ}$, $790^{\circ}$, $1130^{\circ}$, $-70^{\circ}$, $-560^{\circ}$, and $-1010^{\circ}$ that have the same terminal side as $70^{\circ}$.

1. Problem Description

The problem asks to find the angles among $420^{\circ}$, $790^{\circ}$, $1130^{\circ}$, $-70^{\circ}$, $-560^{\circ}$, and $-1010^{\circ}$ that have the same terminal side as $70^{\circ}$.

2. Solution Steps

Two angles have the same terminal side if their difference is a multiple of $360^{\circ}$.

We check each angle:

* $420^{\circ} - 70^{\circ} = 350^{\circ}$. This is not a multiple of $360^{\circ}$.

* $790^{\circ} - 70^{\circ} = 720^{\circ} = 2 \cdot 360^{\circ}$. This is a multiple of $360^{\circ}$.

* $1130^{\circ} - 70^{\circ} = 1060^{\circ}$. Dividing $1060$ by $360$ gives approximately $2.94$, so it is not a multiple of $360^{\circ}$.

* $-70^{\circ} - 70^{\circ} = -140^{\circ}$. This is not a multiple of $360^{\circ}$.

* $-560^{\circ} - 70^{\circ} = -630^{\circ}$. This is not a multiple of $360^{\circ}$.

* $-1010^{\circ} - 70^{\circ} = -1080^{\circ} = -3 \cdot 360^{\circ}$. This is a multiple of $360^{\circ}$.

Therefore, the angles that have the same terminal side as $70^{\circ}$ are $790^{\circ}$ and $-1010^{\circ}$.

3. Final Answer

$790^{\circ}$ and $-1010^{\circ}$