jmc

We consider a hand at the 12 to be $0^{\circ }$. Now we convert the hour and minute hands to a degree measure from $0^{\circ }$ to $360^{\circ }$. If we divide $360^{\circ }$ evenly among 60 minutes, we get that each minute, the minute hand moves $\frac{360^{\circ }}{60}=6^{\circ }$. So if the minute hand is at 48 minutes, it is at $48\cdot 6^{\circ }=288^{\circ }$.

The hour hand is a little trickier. If we divide $360^{\circ }$ evenly among 12 hours, we get that each hour, the hour hand moves $\frac{360^{\circ }}{12}=30^{\circ }$. Note that the hour hand is not at the 2 since it gradually moves toward the 3 throughout the hour. From the 2 toward the 3, the hour hand has moved $\frac{48}{60}=\frac{4}{5}$ of the way. So the degree measure of the hour hand is $2\frac{4}{5}\cdot 30^{\circ }=84^{\circ }$.

To find the smaller angle formed by the two hands, we can find the larger angle $288^{\circ }-84^{\circ }=204^{\circ }$ and subtract from $360^{\circ }$ to get $\boxed{156^{\circ }}$. Or we know that $84^{\circ }$ is coterminal with (ends at the same place as) $84^{\circ }+360^{\circ }=444^{\circ }$. Now we can subtract $444^{\circ }-288^{\circ }=\boxed{156^{\circ }}$ to find the smaller angle.