What are the ranges of triangle angles?

Question

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: $$0\lt \alpha \le 60 \\ 0 \lt \beta \lt 90\\ 60 \le \gamma \lt 180$$

Answer 1

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

Answer 2

The minimum values of $\alpha $ and $\beta $

Suppose that AB line tends to infinity angle $\gamma $ tends to $180^0$ and the other two angles tends to $0$ but they will not be zero.

Therefore $\alpha ,\ beta $ are greater than zero and can be very close to zero but not equal to zero otherwise the figure would not be a triangle.

The maximum value of $\alpha ,\ beta $

Since $\gamma $ cannot be less than $60^0$ sum of angles in a triangle is $180^0$

Consider $\gamma $ to be $90^0$

$\alpha+\beta \le 180^0-90^0$

Now since $\alpha \le \beta \le \gamma $ the maximum values of $\alpha $ and $\beta $ are very close to $90^0$