In $\triangle PQR$, the sum of angles is $180^\circ$. Given $\angle R=70^\circ$, we have $\angle P + \angle Q = 180^\circ - 70^\circ = 110^\circ$. Since PA and QB are angle bisectors, in $\triangle PMQ$, $\angle QPM = \frac{1}{2}\angle P$ and $\angle PQM = \frac{1}{2}\angle Q$. The sum of angles in $\triangle PMQ$ is $180^\circ$, so $\angle PMQ + \angle QPM + \angle PQM = 180^\circ$. This simplifies to $\angle PMQ + \frac{1}{2}(\angle P + \angle Q) = 180^\circ$. Substituting the sum of $\angle P$ and $\angle Q$, we get $\angle PMQ + \frac{1}{2}(110^\circ) = 180^\circ$. Thus, $\angle PMQ + 55^\circ = 180^\circ$, which means $\angle PMQ = 180^\circ - 55^\circ = 125^\circ$. Alternatively, the angle formed by two angle bisectors is given by $90^\circ + \frac{1}{2}\angle R = 90^\circ + \frac{1}{2}(70^\circ) = 90^\circ + 35^\circ = 125^\circ$.