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What are the angles between the three clock hands at any chosen time?
The common version of this question usually concerns only the minute and hour hands. Here times can include seconds, and second/hour and second/minute hand angles also be calculated.
Remember that the angles in question are those of the smaller sector on the circular clock face - the 'internal' angle. So:

  • what is the angle between the minute & hour hands at 09:10:30?
  • and between the second & hour hands, and second & minute hands?
To run a demo, select a time from the hrs, mins & secs below. Scroll through and click the required numbers which will turn blue.
NB:  reset with thereset buttonbutton before choosing a new time.


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Answers
  • The internal angle between minute & hour hands at 09:10:30 is 147.75°.
  • Between the second & hour hands it is 95.25°, and between the second & minute hands it is 117°.
For the demo:
  • The minute/hour hand angle at the chosen time of is .
  • The second/hour hand angle is , and the second/minute hand angle is .
Notes:
  • If you enter 01:05:27 (as listed in puzzle 2 for overlapping hands) the minute/hour angle is shown to be 0.025°, very close to, but not exactly zero.
    See the Notes to puzzle 2 for the explanation and potential errors arising.
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Maths
Method 1 (see Basics):
The equations for the angular differences D are:
  • min/hr hands: (6×mins + secs/10) - (30×hrs + mins/2 + secs/120), from which:
    D = |5.5M - 30H|, where M includes any seconds as a fraction of a minute, as explained in the Basics.
  • sec/hr hands: (6×secs) - (30×hrs + mins/2 + secs/120), from which:
    D = |719S/120 - 30H|, where in this case the hours term H includes any minutes as a fraction.
  • sec/min hands: (6×secs) - (6×mins + secs/10), from which:
    D = |5.9S - 6M|, where in this case M indicates only the whole minutes.
In each case, the internal angle A = D if D <= 180, but 360 - D if D >= 180.
So, for 09:10:30:
  • min/hr hands: D = |5.5×1030/60 - 30×9| = 212.25, so:  A = 360 - 212.25 = 147.75°.
  • sec/hr hands: D = |(719×30)/120 - 30×910/60)| = 95.25, so:  A = 95.25°.
  • sec/min hands: D = |5.9×30 - 6×10| = 117, so:  A = 117°.
Method 2 (see Basics):
First find the total time in seconds passed since midnight: T = (hrs × 3600) + (mins × 60) + secs.
The equations for the angular difference D are then:
  • min/hr hands: cumulative angular difference C is 1/10 - 1/120 degrees for each second passed, from which:
    D = (T × 11/120) mod 360, as explained in the Basics.
  • sec/hr hands: C will be 6 - 1/120 degrees for each second passed, from which:
    D = (T × 719/120) mod 360.
  • sec/min hands: C will be 6 - 1/10 degrees for each second passed, from which:
    D = (T × 5.9) mod 360.
In each case, the internal angle A = D if D <= 180, but 360 - D if D >= 180.
So, for 09:10:30, when T = (9 × 3600) + (10 × 60) + 30 = 33030 secs:
  • min/hr hands: D = (33030 × 11/120)mod360 = 147.75, so:  A = 147.75°.
  • sec/hr hands: D = (33030 × 719/120)mod360 = 264.75, so:  A = 360 - 264.75 = 95.25°.
  • sec/min hands: D = (33030 × 5.9)mod360, = 117, so:  A = 117°.
Whether D is less than 180° or not is easy to decide mentally for either method, but see below to mathematically emulate this.

Extra maths

To calculate whether an angular difference D is less than 180° or not:
  • 1st way:  round (D / 360) gives 0 for D < 180, and 1 for D >= 180.
  • 2nd way:  (D - Dmod180) / 180 also gives 0 for D < 180, and 1 for D >= 180 (and does not involve rounding).
Now, since internal angle A = D if D <= 180, but 360 - D if D >= 180, these may be combined in the form A = |D - (g × 360)|, where the required value of g is 0 or 1 as calculated by either way.

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