Clock Angle Solutions


Consider a clock with a minute and hour hand. If the time reads 10:30 on the clock, where both hands' positions are accurate to within a minute, what is the radian measure of the least positive angle between both hands? What is the measure in degrees?

Tip: try drawing a sketch to help solve the problem.

We can approach this problem in a number of ways. The nice thing about mathematics, which not enough instructors emphasize, is that there is actually a good deal of flexibility and creativity involved when solving problems. My approach may differ than yours, but as long as the logic behind each of your steps is sound, and you arrive at the correct solution, there's nothing wrong with it!

Make sure to pay attention to the directions carefully. We are asked to find the least positive angle between the hour and minute hands. This tells us we need to measure between 6 and 10.5, rather than 10.5 to 6, clockwise. If we look at this specific problem, we can see it's fairly intuitive that 4.5 hour measures separate the hour and minute hands (as we discussed before, the hour hand moves proportionately according to the minutes past the hour).

Sketch of clock showing 10:30 and the angle between hands shaded in

Now we need to determine the ratio of our 4.5 hour measures to the overall amount of hours, which is 12. \[ \dfrac {\frac 92}{12} = \dfrac {9}{24} = \dfrac 38 \] At this point, we have a factor of $2\pi$ radians: $\frac 38 (2\pi)$, which gives us the answer: $\frac {3\pi}{4}$ radians.

Since $\pi = 180\deg$, we can use $\frac {180\deg}{\pi}$ as the conversion factor to convert radians to degrees. In this case, $\frac {3\pi}{4}\left(\frac {180\deg}{\pi}\right) = 135\deg$.

Well that worked out rather cleanly, but what if we had been asked to find the angle between the hands for the time 10:37? The easiest approach that stands out for me is separately finding the angles for the minute and hour hands, respectively. A good interval in which to measure these angles is in $[0, 2\pi)$, where $0$ is at the 12-o'clock position and the angle increases in a clockwise direction. Finding the positive difference between the angles will give us what we're after. Using a sketch will help us visualize these two angles.

Sketch of clock showing both angles overlayed on top of eachother

Given the time, 10:37, we can first start by calculating the angle of the minute hand, which we'll call $A$. This computation is fairly straightforward. We have 37 out of 60 minutes: \[ A = \dfrac {37}{60} \left( \dfrac {2\pi}{1} \right) = \dfrac {37\pi}{30} \simeq 3.8746 \] Now, calculating $B$, we need to factor in the proportional position of the hour hand between 10-o'clock and 11. We simply need to add the ratio of elapsed minutes to total minutes. \[ B = \left( 10 + \dfrac {37}{60} \right) \left( \dfrac {1}{12} \right) \left( \dfrac {2\pi}{1} \right) = \dfrac {\pi}{6} \left( \dfrac {637}{60} \right) \simeq 5.5589 \] Now that we have the measure of both angles, we find their difference to arrive at the final answer: \[ B - A \simeq 5.5589 - 3.8746 \simeq 1.6842 \textrm{ (rad)} \] If we want to convert to degrees: $B - A \left( \frac {180\deg}{\pi} \right) \simeq 96.5\deg$.

If we use our algebra skills, we can try to generalize this technique so we have a useful formula to use to make future computations easier. Let's define the angle subtended by the hour and minute hands as $\theta$, A as our first angle, and B as our second. We'll use $h$ and $m$ for hours and minutes, respectively. \[ \begin{align} A &= \dfrac {m}{60} \left( \dfrac {2\pi}{1} \right) = \dfrac {\pi}{30}(m) \\ B &= \left( h + \dfrac {m}{60} \right) \left( \dfrac {2\pi}{12} \right) = \dfrac {\pi}{6} \left( \dfrac {60h + m}{60} \right) = \dfrac {\pi}{360} \left( 60h + m \right) \end{align} \] Given $A$ and $B$, $\theta$ is defined such that: \[ \begin{align} \theta &= |B - A| \\ &=\left| \dfrac {\pi}{360}(60h + m) - \dfrac {\pi}{30}(m) \right| \\[0.5em] &= \left| \dfrac {(60\pi(h) + \pi(m)) - 12\pi(m)}{360} \right| \\ &= \dfrac {\pi \left| 60h - 11m \right|}{360} \end{align} \] Try substituting some values in for $h$ and $m$. Using 10:37 again, you should get the same result as before, in radians.