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.

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.
