Problem with $\sin 90^\circ =1$ We know 
$$\sin\theta = \frac{\text{perpendicular}}{\text{hypotenuse}}$$
then $\sin 90^\circ = 1$ refers in this case perpendicular = hypotenuse. But, then, the base becomes $0$ according to the Pythagorean triplet law. How is this possible? 
 A: The Pythagorean Theorem asserts that $a^2 + b^2 = c^2$ such that $c$ represents the hypotenuse, namely the side of a triangle that is directly opposite the $90^\circ$ angle. There has to be a $90^\circ$ angle because this theorem only applies to right triangles. If $\sin 90^\circ = 1$, since $$\sin(\cdot) = \frac{\text{opposite}}{\text{hypotenuse}}$$ then we have $\text{opp} = \text{hyp}$. Let $a = \text{opp}$ then we have $$a^2 + b^2 = a^2$$ and therefore $b^2 = 0\Leftrightarrow b = 0$. This means that we don’t have a right triangle because the side $b$ has no length (it has length $0$) and so it does not exist. We simply have a line $a = c$. And since $a/c = 1$ then $\sin 90^\circ = 1$.
Consider for example, $$x^2 + y^2 = z^2.$$ Let $z = y + 1$, then we have $y = \dfrac{x^2 - 1}{2}$. If $x = z$, then $$y = \frac{(y+1)^2 - 1}{2}\Leftrightarrow 1 = \frac{y + 2}{2}$$ and therefore $y = 0$. We can’t escape the fact that $y = 0$ or $b = 0$ and so this is simply just a special case of $\sin\theta$ such that $\theta=90^\circ$.
A: It is a degenerate case or limit case with $$h\geq b\gg a$$

A: Usually, the trigonometric relations, when defined using triangles, are defined only for the angles in the range $0<\theta<π/2$. However, it's still possible to see why $\sin π/2$ is taken to be $1$ from a right triangle. Recall that $$\sin(\cdot)=\frac{\text{opposite}} {\text{hypotenuse}}. $$ But when the angle is $π/2$ then $\text{opposite}=\text{hypotenuse} $, so that the result follows.
