When a particular cable hangs between two poles the shape is a catenary, f(x)=acosh(x/a) where a is the distance from the sag of the cable to the ground. Suppose the poles are 100 feet apart. Find the slope where the cable meets the pole if the sag is 30 feet.

So from f(x) = a cosh(x/a) I got that f(0) = a and f(50)=acosh(50/a) with 50 being the distance from where the sag is 30 feet to the pole.

I need to solve for a somehow with acosh(50/a)-a=30... that being the equation that can tell me how far the sag is from the ground... and then i can plug a into f'(x)=sinh(x/a) and f'(50)=sinh(50/a) to get the derivative at the pole.... but I am stuck solving for a.