Answer ASAP. Please give correct answer. (Involves a bunch of circles and trig identities)
An 80 metre cable is suspended between 2 poles forming a parabola shape (the end-points of the cable are attached to the top of the poles). The vertex of the parabola is 20 metres above ground and both the poles are 50 metres tall. How far apart are the 2 poles?

45.40

Step-by-step explanation:

First of all, the shape of rope is not a parabola but a catenary, and all catenaries are similar, defined by:

y=acoshxa

You just have to figure out where the origin is (see picture). The hight of the lowest point on the rope is 20 and the pole is 50 meters high. So the end point must be a+(50−20) above the x-axis. In other words (d/2,a+30) must be a point on the catenary:

a+30=acoshd2a(1)

The lenght of the catenary is given by the following formula (which can be proved easily):

s=asinhx2a−asinhx1a

where x1,x2 are x-cooridanates of ending points. In our case:

80=2asinhd2a

40=asinhd2a(2)

You have to solve the system of two equations, (1) and (2), with two unknowns (a,d). It's fairly straightforward.

Square (1) and (2) and subtract. You will get:

(a+30)2−402=a2

Calculate a from this equation, replace that value into (1) or (2) to evaluate d.

My calculation:

a=353≈11.67

d=703arccosh257≈45.40

ok then what happened.