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).

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

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