OCD is not an isosceles triangle

You are correct. My apologies for the error in the previous step. $\angle CDA$ is an angle at the circumference, not an angle within $\triangle OCD$. Let's re-solve it correctly. Step 1: Find $\angle AOC$. 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 $\angle CDA = 36^\circ$. $$ \angle AOC = 2 \times \angle CDA = 2 \times 36^\circ = 72^\circ $$ Step 2: Find $\angle COB$. 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 $\boxed{\text{108}^\circ}$. That's 2 down. 3 left today — send the next one.