Can I divide a square using other congruent isosceles triangles than $45-90-45$ ones? For example, can I use $30-120-30$ congruent triangles to completely cover a square without overlapping or extending outside the square?
My single argument so for is that the corner of the square must be composed only of one type of angle(isosceles triangles have two "types" of them: 2 at the base and one at top). That comes from the fact that $2a + b = 180; a, b > 0; a + b < 90 => a > 90$ (impossible). Also the chosen angle that is chosen for completing the corner must divide 90 exactly. Namely $90/a = i$, where $i$ is an positive integer.
It cannot be done with $30^\circ$-$30^\circ$-$120^\circ$ triangles. Suppose the unit square is tiled with $n$ such triangles, whose equal sides have length $s$ and whose longest sides have length $t=\sqrt{3}s.$ The area of each triangle is $$\frac12 s^2\sin{120^\circ}={s^2\sqrt{3}\over4}=\frac1n,$$ so that $$\frac1s={\sqrt{n}\sqrt[4]{3}\over2},$$ an algebraic number of degree $4$.
On the other hand, the tiling divides each side of the square into line segments of lengths $s$ and $t$. Suppose one side is covered by $a$ segments of length $s$ and $b$ segments of length $t$ where $a$ and $b$ are nonnegative integers, not both $0$. Then we have $$as+bt=1\implies as+b\sqrt{3}s=1$$ so that $$\frac1s=a+b\sqrt{3}$$ an algebraic number of degree $2$, contradicting the earlier result.
It should be possible to extend this line of reasoning to solve the general problem, but I don't think I remember enough algebra.