Two vertical poles have heights of 7 feet and 12 feet. A rope is stretched from the top of each pole to the bottom of the other exactly how far above the ground do the ropes cross?

Question

Similarity triangles

And it is a word problem

Answer 1

me make distans tween poles=7 ft =hite av left pole tu make it eezeeer

rope on left side make isosaleez triangel with 2   45 deg angels

tan(45 deg)=1, so rope go down 1 ft for eech ft awae from its pole

other rope...get invers tan 12/7...tan-1(1.71428571428571428...)

Yu get 59.7435628 deg...kloes nuf yu kan kall it 60 deg

2 ropes gotta drop same distans...d1=(7-d1)*1.7142857142857...

d1=7*1.7142857 -d1*1.7142857

2.7142857d1=11.999999999...make that d1=12/2.7142857

=4.421052631579 from left pole (short pole)

Answer 2

The intersection of the ropes is height h above the ground. Let's suppose that a perpendicular from this point is a distance x from the base of pole 1. Then in triangle ABC, h/7=(a-x)/a (similar triangles). In triangle ADC h/12=x/a. So x=ah/12 and h/7=(a-ah/12)a=1-h/12, h(1/7+1/12)=1, 19h=84, h=84/19=4.421 ft.

The picture below contains another solution method.