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.