Albert said:$\triangle ABC ,\,\, given :$
(1) sum of its two angles=$n^o$
(2) if $\alpha$ is the largest angle
and $\gamma $ is the smallest angle ,we have :$\alpha - \gamma =24^o$
please find the range of $n$
it is about any 2 angles, the range of $n$ is $[104^\circ,136^\circ]$I like Serena said:The boundaries are determined by $\beta$ which much be between $\alpha$ and $\gamma$.
In the extreme cases, we have:
$$\alpha = \beta, \gamma=\alpha -24^\circ \qquad\qquad (1)$$
respectively
$$\alpha, \qquad \beta = \gamma=\alpha -24^\circ \qquad (2)$$
Since the sum of the angles must be $180^\circ$, we get:
in case (1): $3\alpha - 24^\circ = 180^\circ \Rightarrow \alpha = \beta = 68^\circ, \gamma=44^\circ$
in case (2): $3\alpha - 48^\circ = 180^\circ \Rightarrow \alpha = 76^\circ, \qquad \beta = \gamma=52^\circ$
It is not clear to me of which two angles $n$ would be the sum.
If it is about $\alpha + \gamma$, the range of $n$ is $[112^\circ, 128^\circ]$.
If it is about any 2 angles, the range of $n$ is $[104^\circ,136^\circ]$.