How would a triangle for sin 90 degree look

One could think as follows: Since the side opposite to the angle B is the hypotenuse, then: $$ sinB = \frac{hypotenuse}{hypotenuse} = 1$$

However, one must always look carefully and rigorously to definitions. The sine of an angle is defined as the division of the opposite side to the angle by the hypotenuse. The hypotenuse can't be one of the sides. Having a look at the trigonometric circle might help:

The sine is defined here always as the value of $y$ divided by 1 (hypotenuse) When the angle is 90°, then $y=1$. I don't want to sound rude, but there is not a lot of sense in "choosing the sides" for calculating the sine of 90°. Hope i've helped.

The short answer is that you can't construct a right angle triangle to visualise $\sin 90^{\circ}$.

The "opposite" (or "perpendicular" as per your nomenclature) side has to be distinct from the hypotenuse, it cannot be the same side. But in a plane figure with two right angles opposite one another, you will never have an apex as the parallel sides won't ever meet (i.e. you can't construct such a triangle).

You can get arbitrarily close by having one true right angle with the other angle approaching very close to a right angle (with the third angle getting progressively smaller). This will result in a very "tall" and "narrow" right triangle if you orient it a certain way. From this you can tell that the side opposite the right angle will become closer and closer to the length of the hypotenuse, with the sine of that angle approaching $1$ from below. In other words, you can construct plane right triangles allowing you to visualise $ \sin x \to 1$ as $x \to 90^{\circ}$, but you can't actually draw the limiting case.

One other way to find angles -

$\sin \theta = \frac{\text{Opposite Side}}{\text{Hyphotenuse}}$

$\cos \theta = \frac{\text{Adjacent Side}}{\text{Hyphotenuse}}$

Edit -

In your case -

$\sin 90° = \frac{\text{Opposite Side}}{\text{Hyphotenuse}}$

And in this case Opposite Side is Hypotenuse itself.