Dodecagon Angles

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 $$\angle OPA = 45^{\circ}$$

$$\angle BOP = 120^{\circ}$$ so $$\angle OPB = 30^{\circ}$$ Therefore $$\angle APB = 45^{\circ} + 30^{\circ} = 75^{\circ}$$ Alternatively, $$\angle AOB = 150^{\circ}$$ 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, $$\angle APB = 75^{\circ}$$