The question asks for the smaller angle between the clock hands at 10 minutes to 2.
Step 1: Determine the exact time.
10 minutes to 2 is equivalent to 1:50.
Step 2: Calculate the position of the minute hand.
The minute hand moves $360^\circ$ in 60 minutes, which is $6^\circ$ per minute ($360^\circ / 60 = 6^\circ$).
At 50 minutes past the hour, the minute hand is at the number 10. Its position from the 12 o'clock position is:
Angle$_{minute} = 50 \text{ minutes} \times 6^\circ/\text{minute} = 300^\circ$
Step 3: Calculate the position of the hour hand.
The hour hand moves $360^\circ$ in 12 hours, which is $30^\circ$ per hour ($360^\circ / 12 = 30^\circ$).
It also moves continuously as the minutes pass. In 60 minutes, it moves $30^\circ$, so it moves $0.5^\circ$ per minute ($30^\circ / 60 = 0.5^\circ$).
At 1:50, the hour hand is past the 1. Its position from the 12 o'clock position is:
Angle$_{hour} = (1 \text{ hour} \times 30^\circ/\text{hour}) + (50 \text{ minutes} \times 0.5^\circ/\text{minute})$
Angle$_{hour} = 30^\circ + 25^\circ = 55^\circ$
Step 4: Calculate the angle between the hands.
The difference between the two angles is:
Angle Difference $= |\text{Angle}_{minute} - \text{Angle}_{hour}|$
Angle Difference $= |300^\circ - 55^\circ| = 245^\circ$
Step 5: Find the smaller angle.
Since the angle between the hands can be measured in two ways (clockwise or counter-clockwise), we take the smaller angle. The sum of the two angles is $360^\circ$.
Smaller Angle $= 360^\circ - \text{Angle Difference}$
Smaller Angle $= 360^\circ - 245^\circ = 115^\circ$
Therefore, the smaller angle between the hands of a clock at 10 minutes to 2 is $115^\circ$.