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
 A: For $(a,b,c)$ to represent angles of a triangle, we should have $a+b+c=180$. Two triangles are equivalent iff their largest angles are the same. For example, $(61,60,59) \sim (61,61,58)$.
Note that at least one of $a,b$ or $c$ must be at least $60$ for the triangle to exist.
So to count the number of equivalence classes, we just have to look at the possible values that the largest angle can take. Suppose $a$ is the largest angle, then $60 \leq a \leq 178$. So the number of equivalence classes will be $119$.
Note that the size of the classes can be different. For example
$$[(60,60,60)]=\{(60,60,60)\} \qquad \text{ and } \qquad [(61,60,59)]=\{(61,60,59), (61,61,58)\}.$$
A: 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.
A: Your question has been answered nicely already but I would like to add a more general approach.

Suppose that $X$ is some set and $R$ is an equivalence relation on $X$ characterized by:$$xRx'\iff f(x)=f(x')$$ where $f$ denotes some function that has set $X$ as its domain.
If $Y:=\{f(x)\mid x\in X\}$ and $E$ denotes the set of equivalence classes then in this situation there is a bijection $E\to Y$ prescribed by: $$[x]\to f(x)$$It is not difficult to prove that the function is well-defined and its inverse is clearly the function $Y\to E$ prescribed by:$$y\mapsto f^{-1}(\{y\})=\{x\in X\mid f(x)=y\}$$
This tells us that the set of equivalence classes has the same cardinality as the image of the function involved.
In your case the function prescribed by: $$(a,b,c)\mapsto\max(\{a,b,c\})$$is characterizing so that finding the number of equivalence classes comes to the same as finding how many values can serve as such maximum value.
