<s>[INST] What are the ranges of triangle angles? Lets say, that $\alpha \le \beta \le \gamma$. As shown here, $60 \le \gamma \lt 180$. What are the minimum and maximum values of $\alpha$ and $\beta$?

The answer:
$$0\lt \alpha \le 60 \\ 0 \lt \beta \lt 90\\ 60 \le \gamma \lt 180$$
 [/INST] If $\alpha \le \beta \le \gamma$ are triangle angles, then their ranges are:
$$
0^\circ \lt \alpha \le 60^\circ\\
0^\circ \lt \beta \lt 90^\circ\\
60^\circ \le \gamma \lt 180^\circ$$
Proof
The lower $\alpha$, $\beta$ and the upper $\gamma$ bounds can be determined from the case, when $\gamma \rightarrow 180^\circ$, $\alpha \rightarrow 0^\circ$, $\beta \rightarrow 0^\circ$. Since $\alpha + \beta + \gamma = 180^\circ$, $\gamma$ cannot be greater than $180^\circ$. And $\alpha$, $\beta$ cannot be $\le 0^\circ$, because when $\alpha = 0^\circ$ we get a line, not a triangle. Hence, $0^\circ < \alpha$, $0^\circ < \beta$ and $\gamma < 180^\circ$.
The upper $\beta$ bound can be determined from the case, when $\gamma \rightarrow 90^\circ$, $\beta \rightarrow 90^\circ$ and $\alpha \rightarrow 0^\circ$. Since $\alpha \ne 0^\circ$, $\beta \le \gamma \lt 90^\circ$. Hence, $\beta \lt 90^\circ$.
Since $\alpha \le \beta \le \gamma$, we get that the upper $\alpha$ bound is when $\alpha = \beta = \gamma = 60^\circ$. So, $\alpha \le 60^\circ$.
MathLove proved that the lower $\gamma$ bound is $60^\circ$. The proof is: 

Let $\alpha \le \beta \le \gamma$ be the inner angles of a triangle. Suppose that $\gamma \lt 60^\circ$. Then, $180^\circ = \alpha + \beta + \gamma \lt 60^\circ + 60^\circ + 60^\circ = 180^\circ$ which is a
  contradiction.

Hence, $60^\circ \le \gamma$. $\Box$
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