Let me add something to Ethan's excellent answer: for the unit circle approach, you have to know some way to say "we've gone $t$ units around the unit circle." That means having a way to measure the length of a curve ... which is typically rigorously defined only when you've done "arclength" in calculus. So there's a little bit of a chicken-and-egg problem here. For angles in the plane (and trig functions associated to them), you can either prove from your axioms that congruent angles have the same "measure" (or assume it, if you're using Hilbert's axioms for plane geometry), or simply assign sine and cosine values to "angles" (however you may have defined those), and then prove that if two angles are congruent, their sines are the same, etc. Most geometry and/or trig books don't bother with this -- they say "we can measure angles with a protractor", and move on. That's probably not a bad idea, but it's not a formal definition of angle measure.
Notice, by the way, that the "right angles" approach and the "circle" approach are, for angles between 0 and 90 degrees, essentially identical. For if you draw a radius from the origin $(0, 0)$ to a point $(x, y)$ of the unit circle with $x, y > 0$, then you can drop a line from $(x, y)$ to the point $(x, 0)$ on the $x$-axis, and then go from there back to the origin. That's a right triangle, with hypotenuse 1 (because it's a unit circle!). The sine is therefore just the length of the vertical segment (which is $y$) divided by the hypotenuse (which is $1$), so the sine is $y$. The cosine is similarly $x$.
The only tricky thing here is that I've defined sine and cosine for the angle $s$ subtended at the origin, but the usual circle definition says "you walk a distance $t$ along the unit circle, counterclockwise starting at $(1,0)$, and get to a point $(x, y)$. We then say that $\cos t = x, \sin t = y$." So the question is "Is the length $t$ of the arc between $(1,0)$ and $(x, y)$ the same as the measure $s$ of the angle subtended at the origin?" The answer's "yes", of course, but it's not a completely trivial statement.