Proof by contradiction using a pigeon hole argument:
Suppose there exists such a triangle, with integral side measures a,b, and c and integral angle A,B, and C,
with angles A,B, and C NOT ALL EQUAL.
sin A/ a = sin B / b = sin C / c by Law of Sines and
These equations can be rewritten as: b = a*sinB/sinA, c = b*sin C / sin B, and a = c * sin A/ sin C
So the sine in the numerator MUST be an integral multiple of sine in the denominator.
So we must find angles A.B and C such that sin B = k * sin A, sin C = n * sin B, and sin A = m * sin C
for integers, k,m, and n.
Then, B = inv_sin(k * sinA) , C = inv_sin( n * sin B) and A = inv_sin(m * sin C)
Plugging the first into the second, C = inv_sin( n * k * sin A)
Plugging that into the third:
A = inv_sin(m * n * k * sin A)
THen, taking the sine of both sides:
sin A = m*n*k * sin A
1 = m*n*k
this forces m=n=k = 1, since m,n, and k are integers, unless two of the three are both -1.
Then, in the former case, the sine functions are all identical, per the statement above in bold.
This forces the angles to be the same, which is a contradiction, since the triangle is given to
be not equilateral.
Now , if any two of the three fixed integers of n,m, and k are -1. then two pair of these trig functions have
OPPOSITE values, since their quotient is -1. In that case, sin T = -sin T.
Then 2*sin T = 0 ---> sin T = 0 ----> T = 0 or 180 which cannot be angle measures of a triangle.
This completes the proof.
A cleaner argument feature law of cosines and sines, stating the cosine and sine of any angle other
than 30 and 60 degrees respectively are irrational. Specifically, sin 30 = cos 60 = sin 150 = 1/2 and
sine and cosine trig functions are irrational everywhere else. A triangle with two angles of degree 30 each, forces the 3rd angle to be 120 which has irrational trig functions. 60 degree angles force the equilateral, and
the 150 degree angle will not fit at all.