What is the greatest area difference between two "nice triangles"? We call a triangle "nice if all angles are between $45$ and $90$ degrees (including $90$ and $45$ itself) and all sides are between $1$ and $2$ (including $1$ and $2$ itself). What is the greatest area difference between two "nice triangles"?
My attempt: Because we have side and angle limits the best way to finding area is using the formula $S=bc\cos{A}$. We should find the greatest and lowest area. But here I got stuck and I don't know how to have both limits with each other. First I thought that the maximum area is $A=90^\circ $ and $b=c=2$. But then I saw that then we have$a=2\sqrt{2}>2$. Could you please give a way?

The Maximal triangle has two sides of length $2.$
for any triangle that had only 2 sides of lenght 2, we could extend the second longest side until it was length 2, and create a larger triangle.
$Area = \frac 12 BC\sin a\\
B,C = 2$
maximize $2\sin a$
constrained by:
$4 \sin \frac {a}{2} \le 2$
$\sin \frac {a}{2}\le \frac 12\\
\frac {a}{2}\le 30\\
a\le 60$
Since $\sin a$ is strictly increasing between 0 and 90.
$a = 60\\
Area = 2\sin 60 = \sqrt 3$
then
minimize $\frac 12 \sin a$
constrained by:
$4 \sin \frac {a}{2} \ge 1$
and again $a = 60$
