In moving from one number on the clock face to the next, a hand moves $360 \div 12 ^\circ = 30 ^\circ$.

At 2.30, the hour hand will be exactly half way between the $2$ and the $3$, and the minute hand will be exactly on the $6$.

So the angle between the two hands will be $3 \times 30^\circ + 15^\circ = 105^\circ$.

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.