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.

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
 * $H$ is the hour.
 * $M$ is the minutes past the hour.
 * $M_{Σ}$ is the number of minutes since 12 o'clock. $$ M_{\Sigma} = (60 \times H + M)$$

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 If the angle is greater than 180 degrees then subtract it from 360 degrees.
 * $H$ is the hour
 * $M$ is the minute

Example 1
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}$$

Example 2
The time is 10:16.


 * $$\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} \ \ ( > 180^{\circ})\\ &= 360^{\circ} - 212^{\circ} \\ &= 148^{\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.$\overline{45}$, 2:10.$\overline{90}$, 3:16.$\overline{36}$, 4:21.$\overline{81}$, 5:27.$\overline{27}$. 6:32.$\overline{72}$, 7:38.$\overline{18}$, 8:43.$\overline{63}$, 9:49.$\overline{09}$, 10:54.$\overline{54}$, and 12:00. (0.$\overline{45}$ minutes are exactly 27.$\overline{27}$ seconds.)