Samiksha Jaiswal (Editor)

Clock angle problem

Updated on
Edit
Like
Comment
Share on FacebookTweet on TwitterShare on LinkedInShare on Reddit
Clock angle problem

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

Contents

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.

Equation for the angle of the hour hand

θ hr = 0.5 ∘ × M Σ = 0.5 ∘ × ( 60 × 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.
  • MΣ is the number of minutes since 12 o'clock. M Σ = ( 60 × H + M )

    • Equation for the angle of the minute hand

    θ min. = 6 ∘ × 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:

    θ hr = 0.5 ∘ × ( 60 × 5 + 24 ) = 162 ∘

    The angle in degrees of the minute hand is:

    θ min. = 6 ∘ × 24 = 144 ∘

    Equation for the angle between the hands

    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 ) |

    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.

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

    Example 2

    The time is 10:16.

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

    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.

    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

    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.)

    References

    Clock angle problem Wikipedia
    (Text) CC BY-SA