# Length between edge of ball supporting a ramp and a wall without similar triangles

1. Sep 2, 2007

### GloryUs

1. The problem statement, all variables and given/known data
A ramp is made by laying one end of a board on the level floor and the other end on top of a 4.0-foot-high wall, which is 8.0 horizontal feet away. A 3.0 foot-in-diameter ball is rolled under the ramp until it is wedged between the underside of the ramp and the floor. How far (horizontally) from the wall is the point where the ball rests on the floor?

***A photo of the diagram is attached***

2. Relevant equations
Any equation may be used except for similar triangles.

The answer (given at the bottom of the worksheet) is 1.6 ft
I have to find out how to get there

3. The attempt at a solution
Every technique I have tried to solve this problem ends up giving me the answer of 2 ft, which is what the answer would be if I used similar triangles.

*When looking at the diagram, it is easy to see that the length in question is just a little longer than 1.5 (the radius of the ball)

*Some attempts I have had are...
-Use the lengths given on the large triangle to create a tangent equation and get the smallest angle measurement (26.5651 degrees), which also provides me with the 3rd angle measurement (63.4349 degrees). Then I used the law of sines to (sin26.5651/3.0 = sin63.4349/x) to get the second-leg-length of the smaller triangle. That equation gave me the answer 6. When subtracted from the longer-leg-length of the larger triangle, the final result is 2, which isn't right.

-Try and incorporate the area of the circle with the area of the triangle with known sides. I couldn't get that to work, I just got the area of the triangle that surrounds the circle.

-Create equations for each triangle and set them equal to each other. After finding the hypotenuse of the large triangle (8.94427 ft), I created the equation 73-16x+x^2 = 8.94 - sq.rt.(x^2+1), and from there I got stuck.

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2. Sep 2, 2007

### learningphysics

Is this a geometry class? What section are you studying... the way I get the answer uses two angles that are congruent... in a way, it is using similar (actually congruent) triangles, so I don't know if it is allowed.

Are there specific theorems that you are allowed to use? Not being able to use similar triangles in any way is hard...

3. Sep 2, 2007

### GloryUs

This problem is from a worksheet the Physics teacher gave everyone to complete over the summer, so there isn't any specific section that he gave us to help with this problem.

When he said we can't use similar triangles, I can only imagine he meant we can't set up a ratio between the sides of one triangle to another to get our answer. I think other aspects of similar triangles can be used though, like the congruent angles.

Did you get the answer 1.6 using congruent angles?

4. Sep 2, 2007

### learningphysics

Yes... specifically if you draw a line from the center of the circle, to that vertex of the triangle (ie where the two tangents to the circle join)... that line bisects the angle... you can use that to find the distance you need...

you know that angle... plus you know the radius of the circle... you can get the distance you need.

5. Sep 2, 2007

### GloryUs

Thank you so much! I really appreciate you helping me out. : )