- Thread starter MegaMoh
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I am assuming you want to measure angles in degrees - not in radians!As simple as the title says it. Is it possible to have a right triangle with 3 integer sides and 3 integer angles(30 degrees, 11 degrees, 79 degrees, etc.)

What will be the unit of measurements for the length of the sides - angstroms, cm, inches, light-years?

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\(\displaystyle (a,b,c)=(m^2-n^2,4mn,m^2+n^2)\) where \(m,n\in\mathbb{N}\) and \(n<m\)

And so, if you are using degrees to measure the angles, it will suffice to find an ordered pair \((m,n)\) such that:

\(\displaystyle \frac{180}{\pi}\arctan\left(\frac{m^2-n^2}{4mn}\right)=k\) where \(k\in\mathbb{N}\) and \(k<90\).

I would likely use a computer to search for a solution.

Shouldn't there be a condition so that \(\displaystyle m, n \neq 0\)?

\(\displaystyle (a,b,c)=(m^2-n^2,4mn,m^2+n^2)\) where \(m,n\in\mathbb{N}\) and \(n<m\)

And so, if you are using degrees to measure the angles, it will suffice to find an ordered pair \((m,n)\) such that:

\(\displaystyle \frac{180}{\pi}\arctan\left(\frac{m^2-n^2}{4mn}\right)=k\) where \(k\in\mathbb{N}\) and \(k<90\).

I would likely use a computer to search for a solution.

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That is implied with \(m,n\in\mathbb{N}\). The natural numbers are positive integers.Shouldn't there be a condition so that \(\displaystyle m, n \neq 0\)?

Oh, sorry. I thought it was \(\mathbb{R}\) somehowThat is implied with \(m,n\in\mathbb{N}\). The natural numbers are positive integers.