An arc is a poor thing to measure. Three straight distances describe it well enough to stake, and all three are the radius multiplied by some function of half the deflection angle.

Long chord: $C = 2R\sin\dfrac{\Delta}{2}$ — C equals two R sine delta over two. $C$ is the straight line from PC to PT in metres, $R$ the radius, $\Delta$ the deflection. Drop a perpendicular from the centre onto that line and you get two identical right triangles, each with the half-angle at the centre and half the chord opposite it — that is where both the 2 and the halving come from.

External distance: $E = R\left(\sec\dfrac{\Delta}{2} - 1\right)$, where $\sec$ is simply $1/\cos$. $E$ is the clearance from the PI in to the middle of the arc. It answers one question: does the curve miss the thing standing at the corner.

Middle ordinate: $M = R\left(1 - \cos\dfrac{\Delta}{2}\right)$. $M$ is the offset from the middle of the long chord out to the middle of the arc. It answers a different question: how far back must a wall, a cut face or a hedge sit for a driver to see round the bend at all.

$E$ and $M$ are the pair that get confused, so fix them by where they are measured from. $E$ starts at the corner, outside the bend, and reaches in. $M$ starts on the chord, inside the bend, and reaches out. On a flat curve they are nearly equal; on a sharp one they are not remotely, and $E$ is always the larger.

Every one of the three takes $\Delta/2$. Feed it the whole deflection and the answer is not slightly wrong — on a 60° bend the chord comes out at 1.7 times its true length. Halve first, then trig.