How many equivalence classes will there be? Consider the subset $T\subseteq \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$ where the three numbers will be the corner angles (in degrees)
of a (real) triangle. For example $(30, 70, 80)\in T$ but
$(10, 30, 50) \not\in T$ (since $10 + 30 + 50 < 180$), and
$(−10, 20, 170) \not\in T$ (since there would not be a negative angle).
We define a relation on $T$ by $(a_1, b_1, c_1)\sim(a_2, b_2, c_2)$ if and only if the triangles that these triples are from have the same largest angle.
This is the exact question from my e-book. And I am literally confused to tackle with this one.
I know it must include something with combinations but in that way their will be a lot of cases. If I am right then yes I can solve this by $a+b+c=180$ (probably $15931$ is answer) *not sure by applying a particular formula here but I know my professor can't gonna allow me to go with this formula because we haven't covered that in our course. So is there any other way to solve this.
I really appreciate you to read this question. It will be very helpful if you will answer this question. Thanks

First, at least one of $(a,b,c)$ is greater than or equal to $60$. If this were not the case, then
$$a+b+c<60+60+60=180$$
(a contradiction). Also, note that $(60,60,60)\in T$. We also know that $(178,1,1)\in T$. Is there any triangle with an angle of $179$ degrees? No, since this would imply that at least one of the other degrees is not a positive integer. It is not too hard to construct triangles with maximum angle $k$ for any integer $60<k<178$. We conclude there are
$$178-60+1=119$$
(the plus $1$ is because we are counting the integers between the bounds inclusively) different equivalence classes.
