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2024-12-05T13:42:38Z

Adding short description: "Mathematical question"

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{{Short description|Mathematical question}}
{{refimprove|date=November 2014}}

[[Image:clockAngles.jpg|thumb|The diagram shows the angles formed by the hands of an analog clock showing a time of 2:20]]
'''Clock angle problems''' are a type of [[mathematical problem]] which involve finding the angle between the hands of an [[analog clock]].

==Math problem==
Clock angle problems relate two different measurements: [[angle]]s and [[time]]. The angle is typically measured in [[degree (angle)|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.<ref>{{cite journal|title=Angles on the Clock Face|author=Elgin, Dave|journal=Mathematics in School|year=2007|volume=36|issue=5|pages=4–5|jstor=30216063|publisher=The Mathematical Association}}</ref>

===Equation for the angle of the hour hand===
:<math>\theta_{\text{hr}} = 0.5^{\circ} \times M_{\Sigma} = 0.5^{\circ} \times (60 \times H + M)</math>

where:
* {{mvar|θ}} is the angle in degrees of the hand measured clockwise from the 12
* {{mvar|H}} is the hour.
* {{mvar|M}} is the minutes past the hour.
* {{math|''M''<sub>Σ</sub>}} is the number of minutes since 12 o'clock. <math> M_{\Sigma} = (60 \times H + M)</math>

===Equation for the angle of the minute hand===
:<math>\theta_{\text{min.}} = 6^{\circ} \times M</math>

where:
* {{mvar|θ}} is the angle in degrees of the hand measured clockwise from the 12 o'clock position.
* {{mvar|M}} is the minute.

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

:<math>\theta_{\text{hr}} = 0.5^{\circ} \times (60 \times 5 + 24) = 162^{\circ}</math>

The angle in degrees of the minute hand is:

:<math>\theta_{\text{min.}} = 6^{\circ} \times 24 = 144^{\circ}</math>

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

:<math>\begin{align}
\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{align}</math>

where
* {{mvar|H}} is the hour
* {{mvar|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.

:<math>\begin{align}
\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{align}</math>

====Example 2====

The time is 10:16.

:<math>\begin{align}
\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{align}</math>

===When are the hour and minute hands of a clock superimposed?===
[[File:clock_angle_problem_graph.svg|thumb|300px|link=http://upload.wikimedia.org/wikipedia/commons/1/1a/Clock_angle_problem_graph.svg|In this graphical solution, ''T'' denotes time in hours; ''P'', hands' positions; and ''θ'', hands' angles in degrees. The red (thick solid) line denotes the hour hand; the blue (thin solid) lines denote the minute hand. Their intersections (red squares) are when they align. Additionally, orange circles (dash-dot line) are when hands are in opposition, and pink triangles (dashed line) are when they are perpendicular. In [http://upload.wikimedia.org/wikipedia/commons/1/1a/Clock_angle_problem_graph.svg the SVG file], hover over the graph to show positions of the hands on a clock face.]]

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

:<math>\begin{align}
\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{align}</math>

{{mvar|H}} is an integer in the range 0–11. This gives times of: 0:00, 1:05.{{overline|45}}, 2:10.{{overline|90}}, 3:16.{{overline|36}}, 4:21.{{overline|81}}, 5:27.{{overline|27}}. 6:32.{{overline|72}}, 7:38.{{overline|18}}, 8:43.{{overline|63}}, 9:49.{{overline|09}},
10:54.{{overline|54}}, and 12:00.
(0.{{overline|45}} minutes are exactly 27.{{overline|27}} seconds.)

==See also==
*[[Clock position]]

==References==
{{reflist}}

==External links==
* 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

[[Category:Mathematics education]]
[[Category:Elementary mathematics]]
[[Category:Elementary geometry]]
[[Category:Mathematical problems]]
[[Category:Clocks]]

Clock angle problem - Revision history

imported>Belbury: Adding short description: "Mathematical question"