I think you are asking how to find the horizontal distance x between two points when a
catenary of a known length 80 units between them dips 20 units at its midpoint.
Certainly it will be less than 80 units. I tried looking for a site with a quick formula or calculator for you, but they generally take the width as given. The appropriate formulas can be found in the Wikipedia page; I haven't taken the time to solve for what you want.
When I got back to look at details, I found that the formulas there are more exactly what you need than I expected.
The specific formulas you need are found in
Wikipedia here:
However, if both ends of the curve (P1 and P2) are at the same level (y1 = y2), it can be shown that
where L is the total length of the curve between P1 and P2 and h is the sag (vertical distance between P1, P2 and the vertex of the curve).
It can also be shown that
and
where H is the horizontal distance between P1 and P2 which are located at the same level (H = x2 − x1).
Your L is 80 and h is 20. Find a, and then put that into the equation for H, which is the width you want. I used Wolfram Alpha to do the calculations.
And the specific numbers in the problem seem designed. What exactly is the context? In what way has it caused "trouble"? And why is it stated so poorly?
You can check your answer by entering it and the other data
here, and checking the height of the lowest point.
You can do something similar
here, by placing points A and B based on your answer, and setting the length s to 80.