Each side of the dodecagon subtends an angle of

30^{\circ}

at the centre of the circumcircle of the figure (the circle which passes through all 12 of its vertices).

Angle AOP = $90^{\circ}$ so OPA = $45^{\circ}$

Angle BOP = $120^{\circ}$ so OPB = $30^{\circ}$

Therefore APB = $45^{\circ}$ + $30^{\circ}$ = $75^{\circ}$

Alternatively, $∠ AOB = 150^{\circ}$ and, as the angle subtended by an arc at the centre of a circle is twice the angle subtended by that arc at a point on the circumference, $∠ APB = 75^{\circ}$ .