Given the diagram, find the measures of angles BOC, BAC, and BDC.

Since O is the center of the circle, OB and OC are radii. Therefore, $OB = OC$.
This means $\triangle BOC$ is an isosceles triangle.

In an isosceles triangle, the angles opposite the equal sides are equal. So, $\angle OBC = \angle OCB$.
Given $\angle OCB = 30^\circ$, we have $\angle OBC = 30^\circ$.

The sum of angles in a triangle is $180^\circ$.

The value is: $\boxed{\text{a) } \angle BOC = 120^\circ}$

$\angle BAC$ is an angle at the circumference subtended by arc BC.
$\angle BOC$ is the central angle subtended by the same arc BC.
The angle at the circumference is half the angle at the center.

Using the value of $\angle BOC$ from part (a):

The value is: $\boxed{\text{b) } \angle BAC = 60^\circ}$

$\angle BDC$ is an angle at the circumference subtended by arc BC.
$\angle BAC$ is also an angle at the circumference subtended by the same arc BC.
Angles subtended by the same arc at the circumference are equal.

Using the value of $\angle BAC$ from part (b):

The value is: $\boxed{\text{c) } \angle BDC = 60^\circ}$