# Limit for right triangles

1. Nov 14, 2014

### Whateverworks

Hello mathematicians!

I've recently completed a trigonometry course online and find the subject to be of great interest.
I find the laws of sine and cosine fascinating and extremely useful and also, of course, Pythagoras theorem is beautiful as well.

Firstly, I claim no superior knowledge here so if I say something wrong please correct me, that is why I am here.

Now to my question. As I have understood it, for one to make use of the laws of sine or cosine one needs to have some values to get started. I have a hard time figuring out if that is correct. I have used the last days trying to solve a problem I have given myself - however I seem lost and that may be because it's unsolvable!

If I know that a+b+c = x, and that a2+b2+c2=y, in a right triangle, what is the value of a, b and c, respectively.

For a specific case lets say, a+b+c = 70, and that a2+b2+c2=1682

Now if this is solvable I request that you do not post the answer by any means, I simply need to know if that can be solved and I will try again.

Thanks a lot,

//WeW.

2. Nov 14, 2014

### Whateverworks

I just realized I posted this in the wrong sub-forum, my bad!

I request that an admin delete this thread. I have created it in the right sub-forum now.

Sorry again.

//WeW

3. Nov 14, 2014

### Staff: Mentor

Without using the fact that we're talking about a right triangle, the two equations above have an infinite number of solutions. You have two equations in three unknowns, so one of the unknowns is unconstrained.

However, since we're talking about a right triangle, we get a third equation -- a2 + b2 = c2, or equivalently $c = \sqrt{a^2 + b^2}$. I am assuming that c represents the length of the hypotenuse. These equations can be used to simplify the system of three variables into a system of two equations involving only a and b, which you can solve for unique solutions for a and b.

4. Nov 14, 2014

### Whateverworks

That is absolutely amazing. Thank you Sir, my motivation for solving this problem is back up! May you have a good Friday evening.