Interesting twist on the Pythagorean theorem.

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I was looking at the formula for isosceles right triangles and wondered if there was a more general one. here it is:

c^2=a^2+b^2
factor a^2
c^2=a^2(1+b^2/a^2)
sqrt
c=a*sqrt(1+b^2/a^2)
simplify
c=a*sqrt(1+(b/a)^2)

It looks a lot neater when you write it out. Basically, an a to b ratio of 1:1 is a*sqrt(2), 1:2 is a*sqrt(5), 1:3 is a*sqrt(10) and so on.
this isn't the most interesting part. If the quantity under the radical is rational then a and b are the sides of a Pythagorean triple. Although this is interesting I have to admit it's pretty much useless.

Actually, there is. The Pythagorean Theorem is a special case of the Law of Cosines. For any triangle:

c^2 = a^2 + b^2 - 2abcosC

where C is the angle opposite of the length c.

When C = 90 degrees, you have a right triangle, and cos C = 0; therefore,

c^2 = a^2 + b^2.

Context. It only applies to right triangles. If what you say applies then tan(A)=b/0. This formula applies in in the same context as the tangent function. Other than the uselessness of it, it's a valid formula. You won't have a triangle with side a=0 ever , except possibly to prove something wrong. Although, I remember that all the way back in algebra I, they made me define the domain of c given a and b. I believe it came out to a+b>c>|a-b|. If a=0 then b>c>b, which should be clearly impossible. no triangle exists with side a=0, so I don't think we have to worry about the fraction.