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]

θ hr = 0.5 ∘ × M Σ = 0.5 ∘ × ( 60 × H + M ) {\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. M Σ = ( 60 × H + M ) {\displaystyle M_{\Sigma }=(60\times H+M)}

θ min. = 6 ∘ × 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 2:20. The angle in degrees of the hour hand is:

θ hr = 0.5 ∘ × ( 60 × 2 + 20 ) = 70 ∘ {\displaystyle \theta _{\text{hr}}=0.5^{\circ }\times (60\times 2+20)=70^{\circ }}

The angle in degrees of the minute hand is:

θ min. = 6 ∘ × 20 = 120 ∘ {\displaystyle \theta _{\text{min.}}=6^{\circ }\times 20=120^{\circ }}

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

Δ θ = | θ hr − θ min. | = | 0.5 ∘ × ( 60 × H + M ) − 6 ∘ × M | = | 0.5 ∘ × ( 60 × H + M ) − 0.5 ∘ × 12 × M | = | 0.5 ∘ × ( 60 × H − 11 × M ) | {\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.

Δ θ = | 0.5 ∘ × ( 60 × 2 − 11 × 20 ) | = | 0.5 ∘ × ( 120 − 220 ) | = 50 ∘ {\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.

Δ θ = | 0.5 ∘ × ( 60 × 10 − 11 × 16 ) | = | 0.5 ∘ × ( 600 − 176 ) | = 212 ∘     ( > 180 ∘ ) = 360 ∘ − 212 ∘ = 148 ∘ {\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.

θ min = θ hr ⇒ 6 ∘ × M = 0.5 ∘ × ( 60 × H + M ) ⇒ 12 × M = 60 × H + M ⇒ 11 × M = 60 × H ⇒ M = 60 11 × H ⇒ M = 5. 45 ¯ × H {\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