Can a Right Triangle be Solved Without Any Given Values?

  • Thread starter Thread starter Whateverworks
  • Start date Start date
  • Tags Tags
    Limit Triangles
Click For Summary
A right triangle cannot be solved without any given values, as it requires specific measurements to apply the laws of sine or cosine effectively. The user posed a question about solving for the sides of a triangle given two equations involving the sides, which initially seemed unsolvable. However, the introduction of the Pythagorean theorem provides a third equation, allowing for a unique solution for the triangle's sides. This clarification restored the user's motivation to tackle the problem. The discussion highlights the importance of having sufficient information to solve geometric problems.
Whateverworks
Messages
19
Reaction score
0
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 let's 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.
 
Mathematics news on Phys.org
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
 
Whateverworks said:
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 let's say, a+b+c = 70, and that a2+b2+c2=1682
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.

Whateverworks said:
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.
 
Mark44 said:
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.

That is absolutely amazing. Thank you Sir, my motivation for solving this problem is back up! May you have a good Friday evening.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 17 ·
Replies
17
Views
11K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 66 ·
3
Replies
66
Views
7K