How is $\sin 90° = 1$ possible? How can two angles of a triangle be equal to $90°$? If two angles were $90°$, this would mean that the two sides would be parallel and the angle of the third side would be equal to 0. Thus, there would be only two vertices and this wouldn't be a triangle at all, ultimately making $\sin 90° = 1$ impossible.
 A: Consider polar coordinates, $(r \cos\theta, r \sin \theta)$ in a unit circle such that $r=1$. Then, we can see that, the mapping in the first quadrant inside the unit circle is just $(\cos \theta, \sin \theta)$. Now, consider a moving point $A$ starting from $\theta=0°$ to $\theta=90°$ on the circumference of the circle. Now, $OA$ is the hypotenuse of the right angle triangle inside the circle. Now, it is clear that at $\theta=90°$, the hypotenuse and the perpendicular are the same line in the $x$-axis (i.e., they coincide). So, $\sin(90°)=\frac{p}{h}=1$. 
As per your argument, the case is two sides coinciding rather than being parallel, because by Pythagoras theorem, we have $h^2=p^2+b^2$, so if $p$ increases then $b$ must decrease, and $h=p$ iff $b=0$.
Where, h  is the radius of the unit circle, p is the perpendicular drawn from (aka height) the point in the circumference to the x-axis, and b is the distance of intersection of p and x-axis from the origin.
A: You can very well admit the concept of flat triangles (coming in two flavors, with a $0°$ and a $180°$ angle).
Anyway, the usual definition of the sine does not involve a triangle. Rather, the trigonometrical circle and the projection of a point at a given angle to the vertical axis. Indisputably, $\sin90°=1$. Simlarly, $\cos90°=0$.
A: A trivial Pythagorean triple, generated by Euclid's formula where $m=n$ such as $(A,B,C)=(0,2,2)$, is a vertical line $(A=0)$ on the $y$ axis. Since $\frac{C}{B}=1, \sin90^\circ=1$.
A: As noted in a comment, this answer to a previous similar question may well help.
To answer your question directly: I offer that mathematicians made $\sin(x)$ “well defined” for values of $x$ that are smaller than $0^{\circ}$ or larger than $90^{\circ}$ (that is, outside the range of values that are valid for an interior angle of a triangle) when it became useful to do so. “Well defined” is a technical term that means, roughly, that everyone agrees on the answer.
As a simple example, imagine an object that is moving in the plane at constant velocity. The velocity can be described as a speed $v$ and an angle $\theta$ measured clockwise from the $x$ axis. Now suppose that we ask the question, “How fast is the object moving in the $y$ direction?” 
If $\theta$ is less than $90^{\circ}$ then trigonometry gives the answer $v \sin(\theta)$. For values of $\theta$ equal or larger than $90^{\circ}$, it’s possible to calculate an answer by deducting an appropriate multiple of $90^{\circ}$ to bring $\theta$ back into the range of valid triangle interior angles. But that’s tedious.
The answer “should” be $v \sin(\theta)$ for any direction $\theta$. To make this answer work, the definition of $\sin$ has to be extended from the range of valid triangle interior angles $(0^{\circ},90^{\circ})$ to the range of valid directions $[0^{\circ},360^{\circ})$. In this extension definition, $\sin(90^{\circ})=1$.
