the point should be located......feet from the smaller pole to use 19 feet of cable?

Question

two vertical poles of length 4 feet and 6 feet stand 13 feet apart. A CABLE reaches from the top of one pole to some point on the ground between the poles and then to the top of the other pole. Where should this point be located to use 19 feet of cable?

Answer 1

The 4' pole A is set into the ground at point A and the 6' pole B is 13' away at point B. The cable is fixed at point P a feet away from A. The 19' cable is attached to the tops of the poles, forming two right-angled triangles.

Using Pythagoras: 4^2+a^2=x^2, where x is the length of cable between P and the top of pole A. Also, 6^2+(13-a)^2=(19-x)^2. So we have x^2-a^2=16 and 36+169-26a+a^2=361-38x+x^2 and x^2-a^2=36+169-26a-361+38x.

x^2-a^2=16=38x-26a-156, so x=(16+156+26a)/38=(172+26a)/38=(86+13a)/19.

Substituting x=(86+13a)/19 in x^2-a^2=16, we have (86+13a)^2/361-a^2=16, which is (86+13a)^2-361a^2=16*361. So, (86+13a-19a)(86+13a+19a)=5776 and (86-6a)(86+32a)=5776.

7396+2236a-192a^2-5776=0⇒1620+2236a-192a^2=0 or 192a^2-2236a-1620=0. This is a quadratic equation:

48a^2-559a-405=0. From this a=12.3301' approx. The negative solution doesn't apply. So the cable needs to be fixed to the ground 12.33' from the 4' pole.

(The length of cable x=12.9627' from the top of the 4' pole to the ground.)