Clock angle problem

The diagram shows the angles formed by the hands of an analog clock showing a time of 2:20

Clock angle problems are a type of mathematical problem which involve finding the angles 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 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

$\theta_{\text{hr}} = 0.5^{\circ} \times M_{\Sigma} = 0.5^{\circ} \times (60 \times H + M)$

where:

• θ is the angle in degrees of the hand measured clockwise from the 12
• M_Σ is the minutes past 12 o'clock.
• H is the hour.
• M is the minutes past the hour.

Equation for the angle of the minute hand

$\theta_{\text{min.}} = 6^{\circ} \times M$

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:

$\theta_{\text{hr}} = 0.5^{\circ} \times (60 \times 5 + 24) = 162^{\circ}$

The angle in degrees of the minute hand is:

$\theta_{\text{min.}} = 6^{\circ} \times 24 = 144^{\circ}$

Equation for the angle between the hands

The angle between the hands can be found using the following formula:

$\begin{align} \Delta\theta &= \vert \theta_{\text{hr}} - \theta_{\text{min.}} \vert \\ &= \vert 0.5^{\circ}\times(60\times H+M) -6^{\circ}\times M \vert \\ &= \vert 0.5^{\circ}\times(60\times H+M) -0.5^{\circ}\times 12 \times M \vert \\ &= \vert 0.5^{\circ}\times(60\times H -11 \times M) \vert \\ \end{align}$

where

• H is the hour
• M is the minute

If the angle is greater than 180 degrees then subtract it from 360 degrees

Example

The time is 2:20.

$\begin{align} \Delta\theta &= \vert 0.5^{\circ} \times (60 \times 2 - 11 \times 20) \vert \\ &= \vert 0.5^{\circ} \times (120 - 220) \vert \\ &= 50^{\circ} \end{align}$

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.

$\begin{align} \theta_{\text{min}} &= \theta_{\text{hr}}\\ \Rightarrow 6^{\circ} \times M &= 0.5^{\circ} \times (60 \times H + M) \\ \Rightarrow 12 \times M &= 60 \times H + M \\ \Rightarrow 11 \times M &= 60 \times H\\ \Rightarrow M &= \frac{60}{11} \times H\\ \Rightarrow M &= 5.\overline{45} \times H \end{align}$

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

• Clock position

References

1. Elgin, Dave (2007). "Angles on the Clock Face". Mathematics in School (The Mathematical Association) 36 (5): 4-5. JSTOR 30216063.

External links

• http://www.delphiforfun.org/Programs/clock_angle.htm
• http://www.ldlewis.com/hospital_clock/ - extensive clock angle analysis
• http://www.jimloy.com/puzz/clock1.htm