A circle of radius $10$ inches has its center at the vertex $C$ of an equilateral triangle $ABC$ and
passes through the other two vertices. The side $AC$ extended through $C$ intersects the circle
at $D$. The number of degrees of angle $ADB$ is:
$\text{(A)} 15 \quad  \text{(B)}30 \quad \text{(C)}60 \quad \text{(D)}90 \quad \text{(E)}120$
$ABC$ is an equilateral triangle, so ∠$C$ must be $60$°. Since $D$ is on the circle and ∠$ADB$ contains arc $AB$, we know that ∠$D$ is $30$° $\implies \fbox{B}$.