$\because PM$垂直平分$AB$，
$\therefore PA=PB$，
$\therefore \angle B=\angle BAP$，
同理：$QC=QA$，
$\therefore \angle C=\angle CAQ$，
$\because \angle BAC=100^{\circ}$，
$\therefore \angle B+\angle C=80^{\circ}$，
$\therefore \angle BAP+\angle CAQ=80^{\circ}$，
$\therefore \angle PAQ=\angle BAC-\angle BAP-\angle CAQ=20^{\circ}$.
故答案为：$20^{\circ}$.

$\because PM$垂直平分$AB$，
$\therefore PA=PB$，
$\therefore \angle B=\angle BAP$，
同理：$QC=QA$，
$\therefore \angle C=\angle CAQ$，
$\because \angle BAC=100^{\circ}$，
$\therefore \angle B+\angle C=80^{\circ}$，
$\therefore \angle BAP+\angle CAQ=80^{\circ}$，
$\therefore \angle PAQ=\angle BAC-\angle BAP-\angle CAQ=20^{\circ}$.
故答案为：$20^{\circ}$.