2012 AMC 10A Problem 4

Both angles have their corner at $B$ and both use the ray $BA$. The problem does not show a picture, so try the two ways the rays can be placed.

1. Ray $BC$ is $24^\circ$ away from ray $BA$, and ray $BD$ is $20^\circ$ away from ray $BA$.
2. Opposite sides of $BA$. The two angles sit next to each other, so $\angle CBD = 24^\circ + 20^\circ = 44^\circ$.
3. Same side of $BA$. Now ray $BD$ lies inside the angle $ABC$, and the angle between $BD$ and $BC$ is what is left over: $\angle CBD = 24^\circ - 20^\circ = 4^\circ$.
4. It cannot be $0^\circ$: that would need $BC$ and $BD$ to be the same ray, but they are different distances ($24^\circ$ and $20^\circ$) from $BA$.

The smallest possible value is $4^\circ$.

The answer is $\boxed{\textbf{(C)}\ 4}$.

When two angles share a corner and one side, their other sides can land on the same side or on opposite sides, so the angle between them is either the difference or the sum. The words "smallest possible" tell you the picture is not fixed. Draw both pictures and compare.

Both angles have vertex $B$ and share ray $BA$. Ray $BC$ makes a $24^\circ$ angle with ray $BA$, and ray $BD$ makes a $20^\circ$ angle with it. There are two possible pictures.

If $C$ and $D$ are on opposite sides of line $BA$, the angles sit next to each other and $\angle CBD = 24^\circ + 20^\circ = 44^\circ$.

If $C$ and $D$ are on the same side of ray $BA$, the angle $ABD$ lies inside angle $ABC$, and $\angle CBD = 24^\circ - 20^\circ = 4^\circ$.

The smaller of the two is $4^\circ$; it cannot be $0$ because the two rays make different angles with $BA$ and so cannot coincide.

The answer is $\boxed{\textbf{(C)}\ 4}$.