Analog Clock Problem 2

• The minute hand moves $1\text{ s}\times \dfrac{1\text{ min}}{60 \text{ s}}\times\dfrac{360^\circ}{60\text{ min}} = 0.1^\circ$
• The hour hand moves $1\text{ s}\times \dfrac{1\text{ min}}{60 \text{ s}}\times \dfrac{1\text{ h}}{60 \text{ min}} \times\dfrac{360^\circ}{12\text{ h}} = 0.008\overline{3}^\circ$

Thus, the angle between the two is $0.1^\circ-0.008\overline{3}^\circ \approx \boxed{0.0917^\circ}$ .

Through computation, the minute-hand moves $0.1°$ every second while the hour-hand moves $0.008\bar{3}°$ every second. At $12$ $:$ $00$ $\text{AM}$ , if the $12$ marking of the clock is $0°$ , the hour-hand would've moved $0.008\bar{3}°$ and the minute-hand would've moved $0.1°$ after a second. Their gap would be $0.1°-0.008\bar{3}°\approx\boxed{0.0917}\space\text{degrees}$ .