I'm so sorry. I didn't read your original question well enough. If I now understand it correctly, you want to know the smaller of the two angles defined by a center and two points on a circle. For instance, if the center is the origin $(0, 0)$ and one point is $(0, 1)$ and the other is $(-1, 0)$, the two angles are 90 degrees and 270 degrees, and you want $90$ degrees (or $\pi/2$ radians).
Let's call the center $(cx, cy)$, the first point $(px, py)$ and the second $(qx, qy)$.
let vx = px - cx; let vy = py - cy;
let wx = qx - cx; let wy = qy - cy;
let dotProd = vx * wx + vy * wy;
let vLen = sqrt(vx * vx + vy * vy);
let wLen = sqrt(wx * wx + wy * wy);
let angle = Acos(dotProd / (vLen * wLen));
The idea is that $v$ represents an arrow (or "vector") pointing from the center to $P$, and $w$ is a vector from the center to $Q$. The cosine of the angle between two vectors is given by
$$
\frac{v \cdot w} {\|v \| \|w \|}
$$
where $\|v \|$ is the length of $v$ (computed using Pythagoras) and where $v \cdot w$ is the "dot product", computed by multiplying $x$-coordinates, multiplying $y$-coordinates, and summing up the result.
So my program above (in some generic imagined language) computes the dot products and the two lengths, does the division, and then takes the inverse-cosine. That produces a number between $0$ and $\pi$ in most languages. You probably need to multiply by $180/\pi$ to get a number from $0$ to $180$.
OK, leaving the material above, I'll now try to address the "I don't know the center" version of the question. First, a picture:
[![Start and end points, and the two possible circle-centers they define][1]][1]
In this picture, $S$ and $E$ are the start and end of your arc. If we draw the segment between them, then draw the perpendicular bisector of that segment (the long vertical line), the circle center must lie somewhere on that bisector, so that it can have the same distance (the radius, $R$) to both $S$ and $E$. But as I've shown here, there are two possible locations, shown as $C_1$ and $C_2$, where this could be. The good news is that this doesn't matter.
Let's call the midpoint of that picture $M$. Then the triangle $CME$ has an angle, $\theta$ that is half of the angle you're looking for. And the sine of that angle is "opposite over hypotenuse", which is just the distance from $E$ to $M$ divided by $R$. So here goes, once again calling the points $p$ and $q$ instead of $E$ and $S#:
function arcMeasure(px, py, qx, qy, R) {
let xDiff = (px - qx) / 2;
let yDiff = (py - qy) / 2;
pmDistance = sqrt( xDiff * xDiff + yDiff * yDiff);
// that's the distance from one end of the arc to the
// center point I've been calling M
let theta = Asin(pmDistance/R); // half the angle we're looking for
return (2 * theta);
}
...and that should do the trick. It'll fail when $R$ is zero, but that case doesn't make sense. Otherwise it should get you what you want.
[1]: https://i.stack.imgur.com/yhToK.jpg
