Darn it.I was hoping to use my isometric ellipse protractor.

 

The scaling in X & Y are not the same

If you look carefully at the values you will see that the drawing is definitely not even close to scale.

 

It is a trick question because they drew the drawing to throw heavy thinkers like us into doing the math.

Make the poles 40m tall and you can see that it is a different problem.
Or make the droop 20m above the ground.
Or make the cable longer than 80m, say 100m.

Yeah nah I'm done.

Let me know when you guys run out of patience...

I don't speak for everyone but I'm past ready

 

Yeah nah I'm done.

I don't speak for everyone but I'm past ready

Spoiler: Here's the answer

The rope forms a "catenary" whose distance from the bottom of its curve to the top of the pole is 40m. This fact is derived by subtracting the 10m distance from the rope's bottom to ground from the 50m height of the poles.

Considering that said catenary is perfectly symmetrical, there's an equal amount of rope on both sides. Since the rope's length is given as 80m, and each half of the rope has a vertical distance of 40m, the only way to comply with those restrictions is if the distance between the poles is exactly zero.

Spoiler: A recommendation

Please stop banging your head...

 

Spoiler: Here's the answer

The rope forms a "catenary" whose distance from the bottom of its curve to the top of the pole is 40m. This fact is derived by subtracting the 10m distance from the rope's bottom to ground from the 50m height of the poles.

Considering that said catenary is perfectly symmetrical, there's an equal amount of rope on both sides. Since the rope's length is given as 80m, and each half of the rope has a vertical distance of 40m, the only way to comply with those restrictions is if the distance between the poles is exactly zero.

Spoiler: A recommendation

Please stop banging your head...