OCD is not an isosceles triangle

The angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle.
$\angle CDA$ is the angle at the circumference subtended by arc $AC$.
$\angle AOC$ is the angle at the center subtended by arc $AC$.
Given $\angle CDA = 36^\circ$.

$$\angle AOC = 2 \times \angle CDA = 2 \times 36^\circ = 72^\circ$$

Since $AB$ is a diameter, $A$, $O$, and $B$ are collinear, forming a straight line.
Therefore, $\angle AOB$ is a straight angle, which means $\angle AOB = 180^\circ$.
We can write $\angle AOB$ as the sum of $\angle AOC$ and $\angle COB$.

$$\angle AOB = \angle AOC + \angle COB$$

Substitute the known values:

$$180^\circ = 72^\circ + \angle COB$$ $$\angle COB = 180^\circ - 72^\circ$$ $$\angle COB = 108^\circ$$

The final answer is $\text{108}^\circ$.