## 4. Angle between the hands

Calculate the size of the smaller angle between the hour and minute hands of a clock at 13 hours, 8 minutes, and 18 seconds. Express the angle in angular minutes.

## Result: $\quad \mathbf{9 3 9}$
## Solution.

Let the initial position of the clock hands be at noon when the hour and minute hands overlap.

The hour hand of the clock "travels" $30^{\circ}$ in one hour.

The hour hand of the clock "travels" $(30: 60)^{\circ}$ in one minute, so in 8 minutes it "travels" $(8 \cdot 30: 60)^{\circ}=4^{\circ}$.

The hour hand of the clock "travels" $(30: 60: 60)^{\circ}=(30: 3600)^{\circ}$ in one second, so in 18 seconds it "travels" $(18 \cdot 30: 3600)^{\circ}=0.15^{\circ}$.

From noon to 13 hours, 8 minutes, and 18 seconds, the hour hand of the clock "travels" a total of

$$
(30+4+0.15)^{\circ}=34.15^{\circ}
$$

The minute hand of the clock "travels" $(360: 60)^{\circ}=6^{\circ}$ in one minute, so in 8 minutes it "travels" $(8 \cdot 6)^{\circ}=$ $48^{\circ}$.

The minute hand of the clock "travels" $(360: 60: 60)^{\circ}=(360: 3600)^{\circ}$ in one second, so in 18 seconds it "travels" $(18 \cdot 360: 3600)^{\circ}=1.8^{\circ}$.

From noon to 13 hours, 8 minutes, and 18 seconds, the minute hand of the clock "travels" a total of $(48+1.8)^{\circ}=$ $49.8^{\circ}$.

The smaller measure of the angle formed by the hour and minute hands of the clock at 13 hours, 8 minutes, and 18 seconds is $(49.8-34.15)^{\circ}=15.65^{\circ}=(15.65 \cdot 60)^{\prime}=939^{\prime}$.