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]
\thetahr=0.5\circ x M\Sigma=0.5\circ x (60 x H+M)
where:
M\Sigma=(60 x H+M)
\thetamin.=6\circ x M
where:
The time is 5:24. The angle in degrees of the hour hand is:
\thetahr=0.5\circ x (60 x 5+24)=162\circ
The angle in degrees of the minute hand is:
\thetamin.=6\circ x 24=144\circ
The angle between the hands can be found using the following formula:
\begin{align} \Delta\theta &=\vert\thetahr-\thetamin.\vert\\ &=\vert0.5\circ x (60 x H+M)-6\circ x M\vert\\ &=\vert0.5\circ x (60 x H+M)-0.5\circ x 12 x M\vert\\ &=\vert0.5\circ x (60 x H-11 x M)\vert\\ \end{align}
where
If the angle is greater than 180 degrees then subtract it from 360 degrees.
The time is 2:20.
\begin{align} \Delta\theta&=\vert0.5\circ x (60 x 2-11 x 20)\vert\\ &=\vert0.5\circ x (120-220)\vert\\ &=50\circ\end{align}
The time is 10:16.
\begin{align} \Delta\theta&=\vert0.5\circ x (60 x 10-11 x 16)\vert\\ &=\vert0.5\circ x (600-176)\vert\\ &=212\circ (>180\circ)\\ &=360\circ-212\circ\\ &=148\circ\end{align}
The hour and minute hands are superimposed only when their angle is the same.
\begin{align} \thetamin&=\thetahr\\ ⇒ 6\circ x M&=0.5\circ x (60 x H+M)\\ ⇒ 12 x M&=60 x H+M\\ ⇒ 11 x M&=60 x H\\ ⇒ M&=
| 60 | |
| 11 |
x H\\ ⇒ M&=5.\overline{45} x H \end{align}
is an integer in the range 0–11. This gives times of: 0:00, 1:05., 2:10., 3:16., 4:21., 5:27.. 6:32., 7:38., 8:43., 9:49., 10:54., and 12:00.(0. minutes are exactly 27. seconds.)