Question

A cable of 80 meters is hanging from the top of two poles that are both 50 meters from the ground.

what is the distance between two poles to one decimal, if center of cables is:-
a)20 meter above the ground.
b) 10 meter above the ground.

Answer 1

Answer: a) 45.4 m                         b) 0m

To see the solution, check the attachment in this answer.

Let me know if you have any doubt in the solution.

Answer 2

Answer:

(a) 45.4m

(b) no solution

Explanation:

Equation of tangent touching the x-axis is:

$y=a*cosh(\frac{x}{a} )$

(a) Top of pole is $(x, 30)$, we get the equation

$a*cosh(\frac{x}{a} )=30+a$

We know half cable length is 40m. Using this value, we get equation,

$a*sinh(\frac{x}{a} )=40$

Dividing both the equation by a,

$cosh(\frac{x}{a} )=\frac{30+a}{a} \\sinh(\frac{x}{a} )=\frac{40}{a}$

Using $cosh^{2}t-sinh^{2}t=1$ ,

We get,

$\frac{(30+a)^{2} }{a^{2} } -(\frac{40}{a} )^{2} =1$

Solving, we get

$a=\frac{35}{3}$

Putting the value,

$\frac{35}{3}sinh(\frac{x}{\frac{35}{3} } ) =40$

$x = (\frac{35}{3}) arcsinh(\frac{120}{35})$

$x = (\frac{35}{3}) ln(\frac{120}{35}+\sqrt{(\frac{120}{35})^{2}+1 } )$

≈ $22.7$

$2x=(\frac{70}{3} )ln(7)$  ≈ $45.4m$

(b) Top of pole is $(x, 40)$, we get the equation

$a*cosh(\frac{x}{a} )=40+a$

We know half cable length is 40m. Using this value, we get equation,

$a*sinh(\frac{x}{a} )=40$

Dividing both the equation by a,

$cosh(\frac{x}{a} )=\frac{40+a}{a} \\sinh(\frac{x}{a} )=\frac{40}{a}$

Using $cosh^{2}t-sinh^{2}t=1$ ,

We get,

$\frac{(40+a)^{2} }{a^{2} } -(\frac{40}{a} )^{2} =1$

Solving, we get

$a=0$

This means that the wire is hanging directly downwards. The cable doubles up upon itself and thus the two poles coincide. This is logically not possible. Thus no solution to this equation.