In the following figure,$\angle DBC = 70^{\circ}$ and ray $BP$ is the bisector of $\angle DBA$. Find $\angle PBC$ and reflex $\angle PBD$.

(N/A) Given that $\angle DBC = 70^{\circ}$ and $ABC$ is a straight line.
Therefore,$\angle DBA + \angle DBC = 180^{\circ}$ (Linear pair axiom).
$\angle DBA + 70^{\circ} = 180^{\circ} \implies \angle DBA = 110^{\circ}$.
Since ray $BP$ is the bisector of $\angle DBA$,we have $\angle ABP = \angle PBD = \frac{1}{2} \times \angle DBA = \frac{1}{2} \times 110^{\circ} = 55^{\circ}$.
Now,$\angle PBC = \angle PBD + \angle DBC = 55^{\circ} + 70^{\circ} = 125^{\circ}$.
Reflex $\angle PBD = 360^{\circ} - \angle PBD = 360^{\circ} - 55^{\circ} = 305^{\circ}$.