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Problems about the Angle between the Hands of a Clock

Date: 09/08/2005 at 09:02:14
From: Roger
Subject: Geometry angles as the relate to hands on a clock.

At 1:45, the angle of the hands of the clock equals 142.5 degrees.  
The hour hand has moved 3/4 of the way between 1 and 2 O'clock. 3/4 
of 30 degrees (the number of degrees between 1 and 2) is 22.5 
degrees.  So you add 22.5 degrees to the number of degrees between 9 
and 1 o'clock which is 120.  Hence the solution is 142.5 degrees.  
The question is: When is the next time on the clock the angles of 
the arms equal 142.5 degrees?

I'm trying to teach this to my 10th grader.  She agrees that the 
minute hand is moving at a faster rate than the hour hand.  I am 
having difficulty teaching her the formula or the concept of an 
equation that equals 142.5 degrees knowing that it happens at least 
once an hour as it relates to time. 

Doing the math we agree that in 1 minute the little hand (hours) 
will move .5 degrees and the big hand (minutes) will move 6 degrees. 
Or, in a minute the hour hand will move 30 degrees and the minute 
hand 360 degrees. It is understanding the relationship of how the two 
movements would equal 142.5 degrees that is the problem.



Date: 09/08/2005 at 11:59:18
From: Doctor Greenie
Subject: Re: Geometry angles as the relate to hands on a clock.

Hi, Roger--

I have always enjoyed problems like this dealing with the angles 
between the hands of a clock.  There are many variations on the method 
for solving these problems.

You have based your analysis on the number of degrees the two hands 
move in 1 minute:

  hour hand: 0.5 degrees
  minute hand: 6 degrees

Note that another obvious approach would be using the number of 
degrees the hands move in an hour:

  hour hand: 30 degrees
  minute hand: 360 degrees

In either of these approaches, the speed of the minute hand is 12 
times the speed of the hour hand (of course!--because the minute hand 
goes around 12 times every time the hour hand goes around once).  A 
third approach simply uses this ratio of the speeds, noting that 
however far the hour hand moves, the minute hand moves 12 times as 
far.

The key with any of these approaches is to think not of the absolute 
angle between the hands but rather the change in the angles between 
1:45 and the next time the angle is 142.5 degrees.

Let's use your approach to solve this particular problem.  At 1:45, 
the minute hand is 142.5 degrees "behind" the hour hand; we want to 
find the time when the minute hand is 142.5 degrees "ahead" of the 
hour hand.  That means we want to know when the minute hand has 
traveled 285 degrees farther than the hour hand (from 142.5 
degrees "behind" to 142.5 degrees "ahead").

If we let "x" be the number of minutes between the two times, then 
we know

  6x = degrees the minute hand has moved
  0.5x = degrees the hour hand has moved

And we need the minute hand to have moved 285 degrees farther than 
the hour hand:

  6x - 0.5x = 285
  5.5x = 285
  11x = 570
  x = 570/11 (minutes)

The next time the hands of the clock will be separated by 142.5 
degrees is 570/11 minutes after 1:45.

We can of course use the second approach using the numbers of degrees 
per hour that the hands move; then we will get an answer in hours 
instead of minutes:

  x = number of hours after 1:45

  360x = degrees minute hand has moved
  30x = degrees hour hand has moved

  360x-30x = 285
  330x = 285
  x = 285/330 = 57/66 = 19/22 (hours)

It is easily verified that 19/22 hours is the same as 570/11 
minutes, so our two answers are equivalent.

Using the third approach, where we simply note that the number of 
degrees the minute hand moves is 12 times the number of degrees the 
hour hand moves, we have

  x = degrees hour hand moves
  12x = degrees minute hand moves

  12x-x = 285
  11x = 285
  x = 285/11 (degrees)

Again, it is easily verified that the hour hand moving 285/11 degrees 
is equivalent to our earlier answers of 570/11 minutes or 19/22 hours.

At 1:45, the minute hand is 142.5 degrees "behind" the hour hand.  
When I first read your problem, I was thinking you wanted to know 
the next time when this would again be the case.  If that had been 
the problem, then we would have wanted to know how long it takes for 
the minute hand to go from 142.5 degrees "behind" to once again 
142.5 degrees "behind".  In this case, we would want the minute hand 
to have traveled a full 360 degrees more than the hour hand.  So, 
using the first method above we would have

  6x - 0.5x = 360
  5.5x = 360
  11x = 720
  x = 720/11 (minutes)

Or using the second method,

  360x - 30x = 360
  330x = 360
  x = 360/330 = 12/11 (hours)

This last answer gives us a shortcut for solving many problems of this 
type.  Note that the calculations do not involve the 142.5 degree 
angle.  These calculations tell us that for ANY angle between the 
hands of the clock, the amount of time between any two successive 
times when the hands make that angle is always 12/11 hours.  This 
makes logical sense.  In any 12-hour period, the minute hand goes 
around 12 times while the hour hand goes around once.  This means the 
minute hand goes around 11 more times, so it passes the hour hand 11 
times.  It also means that in any 12-hour period, the minute hand will 
make any particular angle with the hour hand 11 times.  And since the 
rates of the two hands are constant, the interval between those 
successive times will be 12/11 hours.

How do we use that fact to find the times when the angle between the 
hands is 142.5 degrees?  We know that at 1:45 the minute hand is 
142.5 degrees "behind" the hour hand.  We then know that it is 142.5 
degrees "behind" the hour hand at the following 11 times:

  1:45
  12/11 hours after 1:45
  24/11 hours after 1:45
  36/11 hours after 1:45
  48/11 hours after 1:45
  60/11 hours after 1:45
  72/11 hours after 1:45
  84/11 hours after 1:45
  96/11 hours after 1:45
  108/11 hours after 1:45
  120/11 hours after 1:45

For the times when the minute hand is 142.5 degrees "ahead of" the 
hour hand, we can start by using a trick.  Since the rates of movement 
of the two hands are constant, and since the hands coincide at 12:00, 
the angle between the hands 1 hour 45 minutes after 12:00 is the same 
as the angle between the hands 1 hour 45 minutes before 12:00--which 
is 10:15.  So the minute hand is 142.5 degrees "ahead of" the hour 
hand at the following 11 times:

  10:15
  12/11 hours after 10:15
  24/11 hours after 10:15
  36/11 hours after 10:15
  48/11 hours after 10:15
  60/11 hours after 10:15
  72/11 hours after 10:15
  84/11 hours after 10:15
  96/11 hours after 10:15
  108/11 hours after 10:15
  120/11 hours after 10:15

Note that one of the times in this list corresponds to the answer we 
found earlier: 19/22 hours after 1:45.

I hope all this helps rather than overwhelms....

Please write back if you have any further questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Euclidean/Plane Geometry

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