Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

## Post a reply

Write your message and submit
|
Options

## Topic review (newest first)

bobbym
2012-11-04 00:57:53

Hi;

This was posted in another thread:

Two large towers are erected on a perfectly level piece of ground. We will call them tower A and tower B. Two marks are drawn on the towers near the top. When a line is drawn from the base of tower A to the mark on tower B it is 50 meters long. When a line is drawn from tower B to tower A's mark it is 40 meters long. The two lines intersect between the towers 10 meters above the ground. How far are the towers away from each other?

This a fairly thorny problem and there are nice solutions to it over in that thread. Let's see if we can solve it in Geogebra.

1) Put a point at (0,0). It will be called A.

2) Create a slider and set Min at 0 and Max at 40 with an increment of .01. Right click the slider and uncheck Fixed object.

3) Move the slider to 25 and create a point in the input bar of (a,0). It will be called B. Resize your screen until you see both points.

4) Use the circle with center and radius tool and click A and enter 50.

5) Make another one at B with a radius of 40.

6) Draw a perpendicular line to the x axis through B. This will represent the rightmost tower.

7) Use the intersection tool to find the points of intersection of the smaller circle and the y axis. D and C are then created. Hide C and the smaller circle.

8)Use the intersection tool to find the points of intersection of the larger circle and the perpendicular line. F and E are then created. Hide E and larger circle.

9) Use the move graphics tool and stretch out the x axis until the drawing looks well spaced.

10) Draw line segment BD and AF. You will see b =40 and f = 50 in the algebra pane.

11) Get the point of intersection of BD and AF, it will called G.

12) In the input bar enter y(G) to get just the y value of point G. It will be created in the Number column of the algebra pane labeled g.

13) Right click g and drag it to the drawing and place right under the slider so you can see it easily.

14) In options set rounding to 15 decimal places and move the slider until g is as close to 10 as possible. I got g = 10.00700381196065. That is not bad. Can we do better?

15) Right click your slider and set increment to .0001. I got g = 10.000117605127501.

16) Keep reducing the increment until you get as close to 10 as possible.

With the smallest increment of .00000001, I got g = 10.000000005695588. Now read the value in the slider, it is a = 37.355085329999866. That is the distance between the towers! It agrees well with the exact answer. You should have a drawing that looks something like the one below.

## Board footer

Powered by FluxBB