I can solve any triangle knowing only 2 sides!

[Trurl]

On my site www.constructorscorner.com

 

I have found a way to break any triangle into 2 right triangles and solve it knowing only 2 sides.

 

I know it sounds unlikely and it could be wrong, but if you go to the homepage you will see 2 trig parabola pdf downloads. The ideas need refined to be simpler, but if you read it all the idea is there.

 

You may also be interested in my math section under the title bar --- ideas and gadgets.

 

Trurl

[ydoaPs]

No you didn't. For triangle ABC with side AB=3 and AC=4, what is the length of side BC?

[mooeypoo]

If you know 2 sides you need to know the angle to figure out the third side.

It doesn't have to be a 90 degree triangle, either, but you need these pieces of information.

 

It's called the "Law of Cosines":

[math]c^2=a^2+b^2+2ab \cos(\theta)[/math]

 

You cannot get the third side without having the angle, because there's an infinite amount of triangles that can be created by those two sides.

 

~moo

Edited October 27, 2009 by Cap'n Refsmmat
make this prettier

[Trurl]

But you didn't even look at the PDF's.

 

For an triangle of 3 and 4, I would say split it into 2 right angles. The first with 3 as the hypotenuse. There are only so many size triangles that have 3 as a hypotenuse. And only 2 that I know of that have a side of 4.

 

So take the length of 3 and substitute x/3 = cos and y/3 = sin. Take this trigonometric pythagorean identity and substitute it into a non pythagorean identity and you will have the inverse cosine and it will give you the angle.

 

Subtract the found values of the right triangle of the 3 hypotenuse from 4 and use that subtracted number again substituted into the pythagorean and other identities.

 

I know my work needs polished, but read through it and give it a chance.

 

At the very least show me where I went wrong.

 

The reason it only takes 2 values is because the angles in my problem share a base. By base I mean a common starting point.

 

Please read the whole 2 PDF's. I welcome feedback; If I'm right or wrong.

[BhavinB]

Label the three sides of a triange as A, B, C

 

if I know A=30 and B=8, what is C?

[mooeypoo]

I did read your PDF, trurl, but I have a very specific problem with what you're suggesting.

 

Look at this picture:

 

All three triangles have the same 2 sides (black sides), and yet in all three, the remaining side (the blue one) is different length.

 

According to your method, you will not be able to differentiate between those three triangles. That's a problem.

 

Knowing the two sides of the triangle is only going to work if you have the angle in between them, or the angle anywhere else, perhaps. But knowing only two sides does not give you a finite answer for the third side.

 

It cannot, because as you can see from the 3 examples in the picture, there are INFINITE amount of distinct triangles you can make with those 2 static sides.

 

~moo

[D H]

Label the three sides of a triange as A, B, C

 

if I know A=30 and B=8, what is C?

Trurl: Here's a clue as to how to solve Bhavin's question: Don't bother.

 

Whatever answer you say it is, Bhavin will say: Nope, wrong answer. He will then demonstrate that yours is not the answer he had in mind by producing a triangle in which one side has a length of 30, another has a length of 8, and third side has a length different from what you answered.

[toastywombel]

On my site www.constructorscorner.com

 

I have found a way to break any triangle into 2 right triangles and solve it knowing only 2 sides.

 

I know it sounds unlikely and it could be wrong, but if you go to the homepage you will see 2 trig parabola pdf downloads. The ideas need refined to be simpler, but if you read it all the idea is there.

 

You may also be interested in my math section under the title bar --- ideas and gadgets.

 

Trurl

 

This is not correct I'll explain why, even before reading the PDF. You may know the length of two of three sides, lets call those sides AB and the unknown C. If the angle of the vertex where A and B meet is increased the length of C will be increased. No matter how you change the triangle you are assuming the size of one of the angles.

 

You have to follow the rules of figuring out the size of a triangle, this is trig 101. In order to figure out all the sides and angles of the triangle you must have at least one of the combinations of given data.

SSS(side side side)

SSA (side side angle)

SAA (side angle angle)

ASA (angle side angle)

SAS (side angle side)

 

I actually tried to do the same thing in my 9th grade geometry class, I thought I could solve a triangle with two sides but you can't, just as explained above. My teacher told me how every year she has one kid who tries to do this. It is good to think outside the box, but its good to remember not everything inside the box is wrong.

 

Cool website BTW .

[Trurl]

Thanks toasty,

 

More clearly stated. We know 1 side of a right triangle (angle 90 degrees) and the path of one of the unknown sides of the right triangle.

 

I know it sounds impossible. But the only thing that makes it possible is the path of one unknown side.

 

I am working on a simplified right up, because I know my current right up is hard to follow.

[ydoaPs]

Thanks toasty,

 

More clearly stated. We know 1 side of a right triangle (angle 90 degrees) and the path of one of the unknown sides of the right triangle.

 

I know it sounds impossible. But the only thing that makes it possible is the path of one unknown side.

 

I am working on a simplified right up, because I know my current right up is hard to follow.

 

Hard to follow because it's wrong? Given only two sides, it could be any of an infinite number of triangles.

[mooeypoo]

Thanks toasty,

 

More clearly stated. We know 1 side of a right triangle (angle 90 degrees) and the path of one of the unknown sides of the right triangle.

 

I know it sounds impossible. But the only thing that makes it possible is the path of one unknown side.

 

I am working on a simplified right up, because I know my current right up is hard to follow.

It doesn't sound impossible, it is impossible.

 

Solve this:

Triangle Side A: 10cm

Triangle Side B: 42cm

Triangle Side C: ?

 

So, here's my solution: C=20cm. Draw it out, see I'm right.

Now wait.. I change my mind. I want C to equal 1cm. Draw it out. See it's right. So.. hang on, I am changing my mind again. I actually want the third side to be 4.33cm. Draw it out, see it's right.

 

These above triangles are ALL possible triangles. Your method cannot differentiate these triangles, therefore it is, by necessity, false.

 

And, to be fair, it's the third time anyone makes this (and other) type of claim in the thread. Ignoring this won't make your theory true..

 

~moo

[the tree]

Just one more point: if a suggested algorithm is claimed to be able to do something that is definitely impossible - there is no reason to inspect the algorithm before stating that it cannot work.

[Sisyphus]

If all you have is two sides, there are an infinite number of possibilities for the third side. Specifically, it can be anything greater than the difference between the two sides and less than their sum.

 

For example:

 

Side A = 3

Side B = 4

Side C = unknown

 

Any number greater than 1 and less than 7 will work for side C.

[Trurl]

Again you haven't read the complete problem.

 

 

Yes I know there are infinitely many triangles with 2 given sides. But I am saying on a right triangle there are only so many definite combinations with 2 given sides. And if we know the path of the hypotenuse travels we can substitute values into an equation and find the sides the hypotenuse fits. Yes there are more than one right angle but the one that fits our needs can be found. As for example the angles on our parabola.

 

Go to constructorcorner.com on the home page and view the example and reread the previous 2 PDF's.

[Mr Skeptic]

Are you claiming your method works for all triangles or just right triangles? Because in the first case, you're wrong and in the second case a guy called Pythagoreas figured it out a couple thousand years ago.

 

Edit: Oh, I see. So you have the length of two sides and what was the third bit of information you have?

[Sisyphus]

If you have two sides and a right angle between them, there aren't just "only so many combinations," there is one answer. But then you don't just have two sides, you have side-angle-side, which is all you need for any triangle.

 

If you mean only so many combinations wherein the triangle can be split into two right triangles, then that is false. Any triangle can be split into two right triangles.

[Bignose]

Go to constructorcorner.com on the home page and view the example and reread the previous 2 PDF's.

 

I read some of it. There are some significant errors.

 

The big on that jumps out to me is:

 

how do you plug [math]5x^2 + \frac{1}{x}-125 = 0[/math] into the quadratic equation??? It isn't a quadratic! It is third order! You have to use the cubic equation to get the right answer!

 

http://en.wikipedia.org/wiki/Cubic_function

 

Also, if your "scosine" is based on triangles on a sphere, that doesn't really apply to triangles in a plane, so it doesn't help solve the problem at hand.

Edited October 27, 2009 by Bignose

[toastywombel]

I think it can be concluded that the math introduced by the OP is most certainly flawed.

[insane_alien]

further more, by knowing the 'path' of the third side means you either need to know an angle or a length.

 

then it is trivial to solve.

[DrP]

Yes I know there are infinitely many triangles with 2 given sides. But I am saying on a right triangle there are only so many definite combinations with 2 given sides.

 

 

You have completely changed your tune then! You said 2 sides unknown in the opening post. NOW you are saying you know 2 sides and one angle (the right angle) - as MrS said. Can't believe you do triangle maths without having heard of Pythag's theorum. What School do you go to?!?

 

Even if it isn't a right angle you can solve it with trig. with 2 sides and one angle.

[iNow]

Will a Mod kindly please change the thread title by adding: [ANSWERED: No, You Can't]

[ydoaPs]

Will a Mod kindly please change the thread title by adding: [ANSWERED: No, You Can't]

lol

[Trurl]

You have completely changed your tune then! You said 2 sides unknown in the opening post. NOW you are saying you know 2 sides and one angle (the right angle) - as MrS said. Can't believe you do triangle maths without having heard of Pythag's theorum. What School do you go to?!?

 

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.

 

 

how do you plug into the quadratic equation??? It isn't a quadratic! It is third order! You have to use the cubic equation to get the right answer!

 

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?

 

Will a Mod kindly please change the thread title by adding: [ANSWERED: No, You Can't]

 

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?

[Sisyphus]

I'm looking at the PDFs and I still don't know what you're talking about. How about this:

 

There is a triangle with side A=3 and side B=5. By your method, what is side C?

 

Obviously, you can't answer that given only that information (your original claim). All you can say is 2 < C < 8. So what I'm asking is, what additional information are you using?

Edited October 27, 2009 by Sisyphus

[DrP]

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.

 

 

Sorry - you are right I misunderstood. It still doesn't work though.