Seemingly Simple

1. Mar 16, 2004

Kafter

OK, my current project is one involving some wuite complex acoustics, all of which I'm starting to grasp but am stumped by seemingly simple trigonometry!? Perhaps someone can help me?....

Draw a rectangle (longest sides top and bottom) with a right-angled triangle underneath it (longest side touching the rectangle (same length) and the other side on the right). Now if there were a beam of light, say, originating at the bottom left hand corner of this diagram aimed to the top line and reflected back so that it ends up at the bottom right-hand point, given that the lengths of all sidse of this quadrangle are known, how can one figure out this path length?

At first I thought I needed more information; but because the angle of incidence = the angle of reflection, there can be only one point where the beam is reflected in order for it's origin and ultimate points to be as given, therefore I assume there is a way of calculating where that point is, and using that, the path length required.

This has been boggling mr for a while so any help is greatly appreciated :D

Kafter

2. Mar 16, 2004

So you have something that looks like this?

I'm not sure I understand what you're asking.

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3. Mar 16, 2004

Kafter

close...
________C____________
| |
| |
| |
| |
| |
| |
| |
A |
|
|
|
|
B

Something more like this but with a hypotenuse between A and B. The path I'm trying to figure out is A to B via C when C is only positioned by the laws of reflection and I have no actual figures determining where it is? Does this make more sense? :D

Sorry, the thread doesn't seem to like my picture, I'll try and attach, stand by

4. Mar 16, 2004

Kafter

Ok, quick paint sketch may sort this out....

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5. Mar 16, 2004

Sure, that can be done.

Let's reflect the ending point (call it B, and the beginning point A) about the line x. Now the line connecting A and B is straight, so we have a right triangle. The vertical side has length z + y, and the horizontal side has length x. The Pythagorean Theorem does the rest.

6. Mar 16, 2004

Kafter

forgive me, but I do not understand your solution...the angle at the reflection point (where a and b coincide with line x) is not necessarily 90 degrees as it looks. Are you assuming the the point of reflection is half-way along line x thus enabling a tringle of (z+y, a+b (hypotenuse), x)? Perhaps another sketch may help me?

7. Mar 16, 2004

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8. Mar 16, 2004

Kafter

Thanks so much! I knew the answer was staring at me, but I guess my brain's been fuddled too much to actually see it! lol Thanks again, I see the derivation now - much appreciated :D You truly are the Cookie Monster ;-)

All I gotta do know is get a and b's individual lengths and I'm home free!

Last edited: Mar 16, 2004