Answer: 65

Solution:

Work in the complex plane with $B = 0$ and $A = 1.$ Rotating $z$ about $P$ through angle $\varphi$ counterclockwise gives $P + e^{i\varphi}(z - P).$ So $A' = e^{i\theta},$ and rotating $B$ clockwise through $\theta$ about $A'$ gives $$\begin{aligned} &B' = A' + e^{-i\theta}(0 - A') \\ &= e^{i\theta} - e^{-i\theta}e^{i\theta} \\ &= e^{i\theta} - 1. \end{aligned}$$

Then $$\begin{aligned} &AB'^2 = \left|e^{i\theta} - 2\right|^2 \\ &= (\cos\theta - 2)^2 + \sin^2\theta \\ &= 5 - 4\cos\theta. \end{aligned}$$ Setting this equal to $\left(\frac{4}{3}\right)^2 = \frac{16}{9}$ gives $4\cos\theta = 5 - \frac{16}{9} = \frac{29}{9},$ so $\cos\theta = \frac{29}{36}$ (indeed positive, consistent with $\theta$ acute). Thus $m + n = 29 + 36 = 65.$