In general the secondary student may not ask why is $\sin 90^\circ = 1$ because they can see the answer in the graph of the sine wave. However the students in grade 8 are not familiar with the graph of sine, so how the teacher could find a simple way to convince the student?? All what they have is the sine definition as a ratio between the lenght of the opposite side to the angle and the length of the side opposite to the right angle.
Maybe draw this picture? Make it clear that the green hypothenuse is fixed in length, but the red altitude is growing and approaching that hypothenuse in length as the angle approaches $90^\circ$.
By their definition (using a right triangle), $\sin 90^\circ$ is undefined, since a triangle cannot have two right angles. You need the definition based on a circle ($\sin \theta = y/r$) to have a value for $\sin 90^\circ$.
I would simply define $\sin(x)$ from the unit circle. I would use a right-angled triangle exactly because (as others have point out already) this definition doesn't work for $90^\circ$. Note that is also seem to work well for negative values.
The sine is the ratio of the "opposite" axis (relative to theta, the angle in question) compared to the hypotenuse. The hypotenuse is always longer because it is opposite the right angle.
But as theta approaches 90 degrees, the length of the opposite axis approaches the hypotenuse as limit (in fact, they both approach infinity).
Conversely, as theta approaches 90 degrees, the third angle approaches zero (90 degrees minus theta), and the axis opposite this angle shrinks as theta angle does. T value of the axis approaches zero as the "non-theta" angle approaches zero.
Can't we think to find $\sin 90^\circ $ directly as the ratio between the side opposite to the angle and the side opposite to the right angle ( the basic defenition ) ?? in this privet case the side opposite to the right angle is the same side ! so $\sin 90^\circ = 1$. Notice the diagram.