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Clock angle problem

Mathematical question

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Source updated August 16, 2026

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Clock angle problems are a type of mathematical problem which involve finding the angle between the hands of an analog clock.

01Math 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 2:20. The angle in degrees of the hour hand is:

$\theta _{\text{hr}}=0.5^{\circ }\times (60\times 2+20)=70^{\circ }$

The angle in degrees of the minute hand is:

$\theta _{\text{min.}}=6^{\circ }\times 20=120^{\circ }$

Equation for the angle between the hands

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

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

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.

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

Example 2

The time is 10:16.

${\begin{aligned}\Delta \theta &=\vert 0.5^{\circ }\times (60\times 10-11\times 16)\vert \\&=\vert 0.5^{\circ }\times (600-176)\vert \\&=212^{\circ }\ \ (>180^{\circ })\\&=360^{\circ }-212^{\circ }\\&=148^{\circ }\end{aligned}}$

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.

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

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.)

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Sources and credits

This article is adapted from the Wikipedia article “Clock angle problem”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

Images, from Wikimedia Commons:

ClockAngles.jpg by Nick Connolly at English Wikipedia, Public domain
Clock angle problem graph.svg by cmglee, CC BY-SA 4.0

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