Clock angle problem
Clock angle problems are a type of mathematical problem which involve finding the angle between the hands of an analog clock.
Math problem
Clock angle problems relate two different measurements: angles and time. The angle is typically measured in degrees from the mark of number 12 clockwise. The time is usually based on a 12-hour clock.
A method to solve such problems is to consider the rate of change of the angle in degrees per minute. The hour hand of a normal 12-hour analogue clock turns 360° in 12 hours (720 minutes) or 0.5° per minute. The minute hand rotates through 360° in 60 minutes or 6° per minute.[1]
Equation for the angle of the hour hand
- [math]\displaystyle{ \theta_{\text{hr}} = 0.5^{\circ} \times M_{\Sigma} = 0.5^{\circ} \times (60 \times H + M) }[/math]
where:
- θ is the angle in degrees of the hand measured clockwise from the 12
- H is the hour.
- M is the minutes past the hour.
- MΣ is the number of minutes since 12 o'clock. [math]\displaystyle{ M_{\Sigma} = (60 \times H + M) }[/math]
Equation for the angle of the minute hand
- [math]\displaystyle{ \theta_{\text{min.}} = 6^{\circ} \times M }[/math]
where:
- θ is the angle in degrees of the hand measured clockwise from the 12 o'clock position.
- M is the minute.
Example
The time is 5:24. The angle in degrees of the hour hand is:
- [math]\displaystyle{ \theta_{\text{hr}} = 0.5^{\circ} \times (60 \times 5 + 24) = 162^{\circ} }[/math]
The angle in degrees of the minute hand is:
- [math]\displaystyle{ \theta_{\text{min.}} = 6^{\circ} \times 24 = 144^{\circ} }[/math]
Equation for the angle between the hands
The angle between the hands can be found using the following formula:
- [math]\displaystyle{ \begin{align} \Delta\theta &= \vert \theta_{\text{hr}} - \theta_{\text{min.}} \vert \\ &= \vert 0.5^{\circ}\times(60\times H+M) -6^{\circ}\times M \vert \\ &= \vert 0.5^{\circ}\times(60\times H+M) -0.5^{\circ}\times 12 \times M \vert \\ &= \vert 0.5^{\circ}\times(60\times H -11 \times M) \vert \\ \end{align} }[/math]
where
- H is the hour
- M is the minute
If the angle is greater than 180 degrees then subtract it from 360 degrees.
Example 1
The time is 2:20.
- [math]\displaystyle{ \begin{align} \Delta\theta &= \vert 0.5^{\circ} \times (60 \times 2 - 11 \times 20) \vert \\ &= \vert 0.5^{\circ} \times (120 - 220) \vert \\ &= 50^{\circ} \end{align} }[/math]
Example 2
The time is 10:16.
- [math]\displaystyle{ \begin{align} \Delta\theta &= \vert 0.5^{\circ} \times (60 \times 10 - 11 \times 16) \vert \\ &= \vert 0.5^{\circ} \times (600 - 176) \vert \\ &= 212^{\circ} \ \ ( \gt 180^{\circ})\\ &= 360^{\circ} - 212^{\circ} \\ &= 148^{\circ} \end{align} }[/math]
When are the hour and minute hands of a clock superimposed?
The hour and minute hands are superimposed only when their angle is the same.
- [math]\displaystyle{ \begin{align} \theta_{\text{min}} &= \theta_{\text{hr}}\\ \Rightarrow 6^{\circ} \times M &= 0.5^{\circ} \times (60 \times H + M) \\ \Rightarrow 12 \times M &= 60 \times H + M \\ \Rightarrow 11 \times M &= 60 \times H\\ \Rightarrow M &= \frac{60}{11} \times H\\ \Rightarrow M &= 5.\overline{45} \times H \end{align} }[/math]
H is an integer in the range 0–11. This gives times of: 0:00, 1:05.45, 2:10.90, 3:16.36, 4:21.81, 5:27.27. 6:32.72, 7:38.18, 8:43.63, 9:49.09, 10:54.54, and 12:00. (0.45 minutes are exactly 27.27 seconds.)
See also
References
- ↑ Elgin, Dave (2007). "Angles on the Clock Face". Mathematics in School (The Mathematical Association) 36 (5): 4–5.
External links
- https://web.archive.org/web/20100615083701/http://delphiforfun.org/Programs/clock_angle.htm
- http://www.ldlewis.com/hospital_clock/ - extensive clock angle analysis
- https://web.archive.org/web/20100608044951/http://www.jimloy.com/puzz/clock1.htm
 |  |
