Tricky Trig?

[Cg1]

Hi all,

I wonder if anyone can tell me if the following is possible please?

I know the co-ordinates of the points x, y and z but have no idea how one might work out the distance d or even if such a solution is possible.



[Really, I want to know where the long line cuts the border of the box and the distance d should give me that I think]

This isn't homework - just something which has cropped up and eludes me. It's been an *awfully* long time since I sat in a maths class and I really don't know where to begin. So if anyone can shed any light on the matter - whilst explaining it in the simplest of terms - then I'd be extremely grateful

TIA.

 

[tkhunny]

Are x and y oriented exactly the same in their respective rectangles? It might not matter.
Are the rectangles oriented in any particular way? It might not matter, but could matter substantially.
Are the rectangles the same size? It might not matter.

If x is (a,b), and y is (c,e), you can extrapolate another point, w at (a,e) which is directly below x and z and on the same horizontal line as y. This gives you another right triangle, xwy, that can help solve your problem, noting similarity with triangle xz{the point you seek}. Then, it might be a trigonometry problem so tricky that it's actually a geometry problem.

Have you considered that your point might move all the way to the corner - or even up the right hand side, or, frankly, anywhere around the rectangle?

You should define your problem before seeking the complete solution. Let's see what you get.

 

[Steven G]

You know points x and y. Find the equation of that line. You know the y coordinate of z. Plug that y value into the equation of the line and solve for x. I am assuming that the base of the top figure is horizontal.

 

[Cg1]

Hi tkhunny, thanks very much for your thoughts.

Yes x and y are the centre points of their respective rectangles which are (currently) the same height, width and orientation. And yes, the point where the line leaves the rectangle certainly can change going, as you say, all the way into a corner even.

I think the rectangles may be a red herring though. Really, the problem can be reduced to this:



If I know the co-ordinates of x, y and z (and therefore know the lengths of xz and xy), can the length of d be found? I'm afraid I'm not capable of defining it in any more helpful a way.

You're right of course, a line from x could be extended vertically down to the same level as y (and then the lengths of all three lines could be found) but, would that help to solve d? I really don't know.

 

[Cg1]

Hi Jomo, thanks very much for looking at this.

To answer your last question first yes, the base of the top figure is horizontal.

And yes, I know the co-ordinates of both x and y and indeed of z as it's half the height of the rectangle (which is known) but as I said to tkhunny above, I think the rectangles are actually causing more trouble than they're worth so have removed them for now. But yes, the co-ordinates of z are known.

When you say "Find the equation of that line" and "Plug that y value into the equation of the line and solve for x"... what does any of that mean please? It's been at least 35 years since I was last in a Maths class and when I was, both I and my Maths teacher found the experience equally stressful...

Something has recently occured to me which may help or may be a complete dead-end but... would bisecting the right angle at z be a step forward do you (or anyone else) think? Such as:



Does knowing the length of xz and now all of the angles in both of the small triangles help to find d? Or am I doing more straw-clutching?

 

[Dr.Peterson]

Here's a picture of what you need.

The little triangle "xz_" (I forgot to label the rightmost point) and the big triangle "xwy" are similar, which means that their sides are proportional. So d/D = xz/xw.

Does that give you enough to work from?

 

[Steven G]

Hi Jomo, thanks very much for looking at this.

To answer your last question first yes, the base of the top figure is horizontal.

And yes, I know the co-ordinates of both x and y and indeed of z as it's half the height of the rectangle (which is known) but as I said to tkhunny above, I think the rectangles are actually causing more trouble than they're worth so have removed them for now. But yes, the co-ordinates of z are known.

When you say "Find the equation of that line" and "Plug that y value into the equation of the line and solve for x"... what does any of that mean please? It's been at least 35 years since I was last in a Maths class and when I was, both I and my Maths teacher found the experience equally stressful...

Something has recently occured to me which may help or may be a complete dead-end but... would bisecting the right angle at z be a step forward do you (or anyone else) think? Such as:

View attachment 11424

Does knowing the length of xz and now all of the angles in both of the small triangles help to find d? Or am I doing more straw-clutching?

There are no triangles nor squares that you must consider to do this problem. As Dr Peterson noted you can use similar or you can try what I posted. Please try both of them!

 

[Cg1]

Hi,

Thanks both for your thoughts - the clue from Dr.Peterson about them being proportional worked a treat! It really is greatly appreciated.

Best wishes,