Clock angle problem

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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.

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]

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

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. ${\displaystyle M_{\Sigma }=(60\times H+M)}$

${\displaystyle \theta _{\text{min.}}=6^{\circ }\times M}$

where:

• θ is the angle in degrees of the hand measured clockwise from the 12 o'clock position.
• M is the minute.

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

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

The angle in degrees of the minute hand is:

${\displaystyle \theta _{\text{min.}}=6^{\circ }\times 24=144^{\circ }}$

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

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

where

• H is the hour
• M is the minute

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

The time is 2:20.

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

The time is 10:16.

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

The hour and minute hands are superimposed only when their angle is the same.

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

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.)

• Clock position

• 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