Did you know that a right triangle can be found using only one side?

[Crazy Horse 11]

It is a more complicated way to write the usual equations.

You can replace "5" with "545349-545344". Nobody does, as it does not add anything new.

You don't have to, but you have the freedom to do so.
Your "method" is just one possible way to pick some specific triangle from the set of all triangles (with that hypotenuse). You can do the same by choosing a, or h, or anything else in the triangle.

A line makes up one side of the triangle what are the other two sides? boo who I don't have enough information?

okay x = 3 and it is 1/5th of the line what are the other two sides? boo who I don't have enough information?

but yes you do!
c = 5x
c = 5 * 3
c = 15
y = 15 - 3
y = 12
c = x + y
15 = 3 + 12
c^2 = cx + cy
side a = sqrt cx = sqrt 45
side b = sqrt cy = sqrt 180
side c = 15

double checking with Pythagoras:
c^2 = a^2 + b^2
225 = 45 + 180 yes the triangle that I just designed is a right triangle using my own methods. But using my methods at least we now know what the right triangle looks like.

What is your answer to the question? Personally I feel that you still don't have enough information if all you are limited to Pythagoras theorem or the law of sine's!

 

[micromass]

I don't get this. If you are given one side, then there are infinitely many right triangles which have that as a side. So why would you expect to get only one solution? Why would we care about a formula that gives only one solution?

 

[Crazy Horse 11]

What you take for granted is that you already have a right triangle to begin with for your exercises.

What you cannot do is calculate a triangle from a single side. Thank-you for allowing me to show you how I can do that using my own methods at that.

 

[Borek]

What you cannot do is calculate a triangle from a single side. Thank-you for allowing me to show you how I can do that using my own methods at that.

OK, so if you are right, you will have no problem solving my problem.

I have a triangle with hypotenuse length equal to 10. Give me the other two sides of the triangle I have on mind.

 

[Crazy Horse 11]

I don't get this. If you are given one side, then there are infinitely many right triangles which have that as a side. So why would you expect to get only one solution? Why would we care about a formula that gives only one solution?

Those are two good questions to ask of yourself from for the answer.

 

[Crazy Horse 11]

OK, so if you are right, you will have no problem solving my problem.

I have a triangle with hypotenuse length equal to 10. Give me the other two sides of the triangle I have on mind.

Frankly how do you know that your other two sides are correct? Show me your design work since you have admitted having a solution on your mind. Did you use any of my methods or not?.

 

[Crazy Horse 11]

Frankly how do you know that your other two sides are correct? Show me your design work since you have admitted having a solution on your mind. Did you use any of my methods or not?.

Let me ask you this instead. At what point do you want your hypotenuse separated into x and y?

 

[AkInfinity]

I cannot believe people are giving you so much greif for this Crazy Horse. As what you found (while not a new rule) is a simple beautiful association.

The mathematical law is of course priceless (although already discovred) but great credit to you for rediscovering.

Here I shall explain. If you are given a hypotenus, and you "assume that you have a right angle" (this part is key) these are all the clues you need for a variant of the proof tequnique known as (Side Angle Side/ Side Angle Angle). Its obvious that for a given hypotenus there can only be one right side angle with respect to the hypotenus.

Briefely put the tecnique invovles alwasy constructing a trialnge of of a line (assuming its the hypotenus) with an angle of 90 degrees. There is only one way this can possibly be done which is why crazy horse alwasy gets the right results.

I would consider this more a tecnique (an extremely usefull and overlooked); but it really isn't a new way or another proof etc; just a very intelligent and well posed method by carefully using the asumptions you can give a problem (the 90 degree ange) and hypotenus length to incredibly simplify a problem; basically what heavyside did to maxwell.

Great Job Crazy Horse if you have any other great ideas let me know I always give good open minded scientific feedback :D

 

[mfb]

okay x = 3 and it is 1/5th of the line what are the other two sides? boo who I don't have enough information?

but yes you do!

Yes of course you can calculate it, and that is nothing new.

What is your answer to the question?

c=15
a=sqrt(cx)=sqrt(45)
b=sqrt(c(c-x))=sqrt(180)

Personally I feel that you still don't have enough information if all you are limited to Pythagoras theorem or the law of sine's!

You can apply Pythagoras twice, or use a formula which is the result of that application. It is called "Kathetensatz" (literally: "theorem of the cathetus") in German, I don't know how it is called in English.

 

[Fredrik]

Closed for moderation.

 

[Borek]

Frankly how do you know that your other two sides are correct?

Once we finish I will gladly give you my result and you will be able to check by yourself, using Pythagorean theorem, that they are correct.

Show me your design work since you have admitted having a solution on your mind. Did you use any of my methods or not?.

I am not going to show them - you said you can calculate just by knowing hypotenuse, it is time that you prove you can do it.

If you need more information, it means you can't solve the problem.