# 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

$\theta_{\text{hr}} = \frac{1}{2}M_\Sigma = \frac{1}{2}(60H + M)$

where:

• $\scriptstyle\theta$ is the angle in degrees of the hand measured clockwise from the 12
• $\scriptstyle H$ is the hour.
• $\scriptstyle M$ is the minutes past the hour.
• $\scriptstyle M_\Sigma$ is the minutes past 12 o'clock.

### Equation for the angle of the minute hand

$\theta_{\text{min.}} = 6M$

where:

• $\scriptstyle\theta$ is the angle in degrees of the hand measured clockwise from the 12 o'clock position.
• $\scriptstyle M$ is the minute.

#### Example

The time is 5:24. The angle in degrees of the hour hand is:

$\theta_{\text{hr}} = \frac{1}{2}(60 \times 5 + 24) = 162$

The angle in degrees of the minute hand is:

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

### Equation for the angle between the hands

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

\begin{align} \Delta\theta &= \left|\theta_{\text{hr}} - \theta_{\text{min.}}\right| \\ &= \left|\frac{1}{2}(60H + M) - 6M\right|\\ &= \left|\frac{1}{2}(60H - 11M)\right| \end{align}

where

• $\scriptstyle H$ is the hour
• $\scriptstyle M$ is the minute

#### Example

The time is 2:20.

\begin{align} \Delta\theta &= \left|\frac{1}{2}(60 \times 2 - 11 \times 20)\right|\\ &= \left|\frac{1}{2}(120 - 220)\right|\\ &= 50 \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{hr}} &= \theta_{\text{min.}}\\ \Rightarrow \frac{1}{2}(60H + M) &= 6M\\ \Rightarrow 11M &= 60H\\ \Rightarrow M &= \frac{60}{11}H\\ \Rightarrow M &= 5.\overline{45}H \end{align}

$\scriptstyle H$ is an integer in the range 0–11. This gives times of: 0:00, 1:05.45, 2:10.90, 3:16.36, etc. (0.45 minutes are exactly 27.27 seconds.)