Two electric poles are placed into the ground. One is 50 feet tall and the other is 30 feet tall. They are 40 feet apart on the ground. You connect the top of each pole to the bottom of the other pole with a wire. The wires need to cross at 15 feet above the ground. At what height do they cross right now? Is it above 15 feet or below 15 feet? Would you move the poles closer together or farther apart to make the wires cross at 15 feet. Explain in detail.
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Geometry: 2 poles 40 ft apart, 50 ft, 30 ft tall....
- Thread starter Ann2343
- Start date
Did you draw a sketch? I did. I saw two right triangles, each with a similar triangle within. I labeled the distance from where the lines cross to the ground, x. I then formed two proportions, one from each of the larger triangles. Solving simultaneously, I got the present distance from the point of crossing to the ground as being 16.25 ft.
If you need more help, please show us what you have done so far.
If you need more help, please show us what you have done so far.
Yes I do need more help. I also drew a sketch. I see the two large right triangles and the two smaller right triangles within the large ones. Im confused as to how you formed a proportion of the triangles. Maybe if you could tell me more of that I would understand it better.Thanks so much.
If it weren't so cumbersome to upload an image I would do so. Since I don't know how, consider one of the triangles. The vertical leg of the right triangle is 50, the horizontal leg is 40. The segment dropped from the intersection of the wires, we will call x. Let's call the intersection of x and the horizontal leg point A. From point A to the vertex of the right angle is y and from point A to the base of the other pole is 40-y. The proportion is...
\(\displaystyle \frac{40-y}{x}=\frac{40}{50}\)
In like manner you can build a proportion involving the other large right triangle. That will give you two equations in two unknowns. Solve for x and you are off to the races. Good luck.
\(\displaystyle \frac{40-y}{x}=\frac{40}{50}\)
In like manner you can build a proportion involving the other large right triangle. That will give you two equations in two unknowns. Solve for x and you are off to the races. Good luck.