Fuzzwood Posted October 27, 2009 Share Posted October 27, 2009 Trurl said: No I haven't changed my original idea. You must first solve 2 right triangles to find a triangle with only 2 sides. It is in my 3rd PDF. Thanks BigNose. At least you looked at the problem. I'm not familiar with the cubic equation. I thought I might be breaking a rule to use the quadratic equation. I meantioned it in the second PDF. Do you see any merit if the correct (cubic) equation was used? Change it to it just might be possible. I should have realized my title would get all the mathematicians fired up. Forget about solving the right triangle with the hypotenuse known and the path known for a moment. If you knew what 2 right triangles lie against a triangle with 2 known sides could you solve the unknown triangle knowing all values for the 2 right triangles? PROVE it works, and not by using a drawing in which you already assume the angle between A and B, your drawing itself shows you need that angle in order to get d and e Link to comment Share on other sites More sharing options...

John Cuthber Posted October 27, 2009 Share Posted October 27, 2009 Since it has already been proved that it cannot work the only outcome of challenging him to "prove it works" will be that he wastes more time and bandwidth. Trurl I can construct two totally different triangles with two sides of lengths 3 and 5. There isn't a unique solution so you cannot sensibly claim to have found "it". I'm beginning to wonder if Trurl can't spell "troll". Link to comment Share on other sites More sharing options...

mooeypoo Posted October 27, 2009 Share Posted October 27, 2009 Trurl said: No I haven't changed my original idea. You must first solve 2 right triangles to find a triangle with only 2 sides. It is in my 3rd PDF. Seeing as errors were found and demonstrated with this idea, you're either incredibly stubborn or you didn't read the responses you recieved. Either way, your idea is clearly not working, and you should revise it, at the very least. Quote Thanks BigNose. At least you looked at the problem. I'm not familiar with the cubic equation. I thought I might be breaking a rule to use the quadratic equation. I meantioned it in the second PDF. Do you see any merit if the correct (cubic) equation was used? Stop ignoring responses. We ALL looked at the problem, and we ALL showed you EXACTLY why your idea is not working. Ignoring us won't make us wrong. It definitely won't make you right. Quote Change it to it just might be possible. I should have realized my title would get all the mathematicians fired up. Forget about solving the right triangle with the hypotenuse known and the path known for a moment. If you knew what 2 right triangles lie against a triangle with 2 known sides could you solve the unknown triangle knowing all values for the 2 right triangles? Stop beating around the bush. Side A: 12 cm. Side B: 10 cm. Solve for side C definitively. Show the above is definitive and that ONLY YOUR SOLUTION is the right one, and you will convince us. Otherwise, your idea is bunk, and if you continue ignoring responses, there won't be any more use in continuing the discussion. ~moo Link to comment Share on other sites More sharing options...

Bignose Posted October 27, 2009 Share Posted October 27, 2009 Trurl said: Thanks BigNose. At least you looked at the problem. I'm not familiar with the cubic equation. I thought I might be breaking a rule to use the quadratic equation. I meantioned it in the second PDF. Do you see any merit if the correct (cubic) equation was used? No, there are other math errors. You didn't even address my concern over your use of the "scosine". The math you wrote up is very unclear. It it very hard to tell how one equation becomes another. And you cannot just "use" equations that don't fit the form. It's not just "breaking a rule". It is using something completely inapplicable to the situation at hand. It is like using a hammer to tighten a nut on a bolt. Or using a tape measure to measure how long it took you to read this post. It is simply a tool that cannot be used to perform the task you need to perform. To use a certain equation, what you have has to fit exactly the conditions to use the equation. And, just because I picked out one math error, doesn't mean that I don't agree with the others brought up in this thread. Please prove your method, especially by answering the direct questions. If you can actually do it, then you will win over converts. Otherwise, it is nothing but words. So, why don't you actually do it? Link to comment Share on other sites More sharing options...

Trurl Posted October 27, 2009 Author Share Posted October 27, 2009 I did not mean to apply you agree with me BigNose. I just meant you actually read the PDF. The Scosine I am referring to is the length a radius must have to reach the same distance as the same radius at a different angle. That is the "path" I am referring to. My math builds on older attempts. Here is the Scosine And the equation I modified Here Here is all my math Though my attempt is far from perfect. I believe there is some merit here. Merged post follows: Consecutive posts mergedBut your right I need more than an concept, I need to mathematically prove it. Link to comment Share on other sites More sharing options...

Klaynos Posted October 28, 2009 Share Posted October 28, 2009 Trurl said: But your right I need more than an concept, I need to mathematically prove it. You can't, the proof that there are an infinite number of triangles with sides a, and b, is trivial and shown in this thread... Link to comment Share on other sites More sharing options...

Mr Skeptic Posted October 28, 2009 Share Posted October 28, 2009 Klaynos said: You can't, the proof that there are an infinite number of triangles with sides a, and b, is trivial and shown in this thread... Right, you might as well be claiming you can solve for a triangle knowing no sides and no angles. Link to comment Share on other sites More sharing options...

toastywombel Posted October 28, 2009 Share Posted October 28, 2009 Trurl said: I did not mean to apply you agree with me BigNose. I just meant you actually read the PDF. The Scosine I am referring to is the length a radius must have to reach the same distance as the same radius at a different angle. That is the "path" I am referring to. My math builds on older attempts. Here is the Scosine And the equation I modified Here Here is all my math Though my attempt is far from perfect. I believe there is some merit here. Merged post follows: Consecutive posts mergedBut your right I need more than an concept, I need to mathematically prove it. Thats like me saying look I solved 1/0. Then when everyone says I am wrong, I respond saying, well I know, but at least I deserve some merit because I tried. Link to comment Share on other sites More sharing options...

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