Analog Clock Problem

The hour-hand moves with a rate $\omega_h = \dfrac {30^\circ}{60} = 0.5^\circ \text{/min}$ and the minute-hand moves with, $\omega_m = \dfrac {360^\circ}{60} = 6^\circ \text{/min}$ .

Let the angle at the marking 12 be $0^\circ$ . Then the angular position of the hour-hand after time $t$ minutes is $\theta_h = \omega_h t = 0.5 t$ in degrees and that of minute-hand is $\theta_m = 6t$ . Since the minute-hand moves faster, the angle between the two hands is given by $\theta_m - \theta_h$ . For the angle to be $180^\circ$ , we have:

$\begin{aligned} \theta_m (t) - \theta_h (t) & = 180 \\ 6t - 0.5t & = 180 \\ t & = \frac {180}{5.5} \approx 32.727 \text{ min} \end{aligned}$

Therefore, Nelson has to wait 32 to 33 minutes .

The formula for getting the angle between the hour hand and the minute hand of an analog clock is $|0.5° \times (60 \times H - 11 \times M)|$ , with $H$ being the hour and $M$ being the minute. In this case, we're trying to find $M$ .

Knowing that minutes should be less than $60$ , when we solve this using calculator, $M$ turns out to be $32.\overline{72}$ ( $98.\overline{18}$ also comes out as an answer but as said before $M$ should be less than $60$ ).

So Nelson would have to wait $\boxed{32 \space \text{to} \space 33 \space \text{minutes}}$ to see the hands to be $180°$ apart.

12:32PM

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