Solution:

A chord subtending a $\theta$-degree arc in a circle of radius $R$ has length $2R\sin\frac{\theta}{2}.$ Write $t = \frac{d}{2},$ so the three chords are $2R\sin t = 22,$ $2R \sin 2t,$ and $2R \sin 3t.$ Using $\sin 2t = 2 \sin t \cos t$ and $\sin 3t = \sin t\,(4\cos^2 t - 1),$ the chords of the $2d$- and $3d$-degree arcs are $22 \cdot 2\cos t = 44 \cos t$ and $22\,(4\cos^2 t - 1).$

The condition "the $2d$-chord is $20$ longer than the $3d$-chord" becomes $44\cos t = 22\,(4\cos^2 t - 1) + 20,$ which simplifies to $44 \cos^2 t - 22 \cos t - 1 = 0.$ From this, $4\cos^2 t = 2\cos t + \frac{1}{11},$ so the $3d$-chord equals $22\left(2\cos t + \frac{1}{11} - 1\right) = 44 \cos t - 20.$

Solving the quadratic, $\cos t = \frac{22 + \sqrt{484 + 176}}{88} = \frac{11 + \sqrt{165}}{44}$ (the $+$ root since $d \lt 120$ means $t \lt 60^\circ,$ so $\cos t \gt \frac{1}{2}$). Then the $3d$-chord is $44\cos t - 20 = 11 + \sqrt{165} - 20 = -9 + \sqrt{165},$ giving $m + n = 9 + 165 = 174.$