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Pythagorean Theorem and non-Right Triangles


Date: 03/09/2002 at 13:16:54
From: Katy Palmen
Subject: Pythagorean Theorem

Why doesn't the Pythagorean Theorem work for triangles other than 
right triangles?

I can provide counter examples to show that it doesn't work for 
specific triangles, such as equilateral or obtuse, but I can't figure 
out a general explanation or proof.

Thank you for your assistance!


Date: 03/09/2002 at 22:34:40
From: Doctor Jeremiah
Subject: Re: Pythagorean Theorem

Hi Katy,

The Pythagorean formula is a special case of a more general equation.  
The full equation is the Cosine Law:

           C^2 = A^2 + B^2 - 2AB cos(c)

You will notice that this equation only degenerates into the 
Pythagorean formula when cos(c) is equal to zero. And that can only 
happen when angle c is 90 degrees. That extra term is the secret to 
why it only works for 90-degree angles.

It all goes back to the geometrical representation of the Pythagorean 
theorem, which you can find in the Dr. Math FAQ:

   http://mathforum.org/dr.math/faq/faq.pythagorean.html   

If the angle is not 90 degrees, you end up with something where the 
two squares don't add up to the third:

 

For example, 3 squared plus 4 squared equals 5 squared, and the sides 
of the triangle are 3, 4, and 5 when the angle is 90 degrees, but 
if you make the angle 77.36 degrees then the side lengths become 4, 4, 
and 5. And 4 squared plus 4 squared does not equal 5 squared.

You can cut out some squares of paper of different sizes: 3x3, 4x4, 
5x5, 6x6, 7x7, 8x8, and see which ones work for right-angled 
triangles.

However these numbers are right and the cosine law can figure out the 
long side:

           C^2 = A^2 + B^2 - 2AB cos(c)
           C^2 = 4^2 + 4^2 - 2(4)(4) cos(77.36)
           C^2 = 16 + 16 - 32 cos(77.36)
           C^2 = 32 - 32 cos(77.36)
           C^2 = 32 - 7
           C^2 = 25
             C = 5

So we can use the cosine law to figure out what the angle is:

           C^2 = A^2 + B^2 - 2AB cos(c)
           5^2 = 3^2 + 4^2 - 2(3)(4) cos(c)
            25 = 9 + 16 - 24 cos(c)
   25 - 9 - 16 = -24 cos(c)
             0 = -24 cos(c)
             0 = cos(c)
     arccos(0) = c
            90 = c

Here is an entry from the archives that gives more examples:

   Basics of Trigonometry
   http://mathforum.org/dr.math/problems/coley.12.12.01.html   

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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