How Many Times The Clock Hands Make an angle Theta? You might have encountered such a question where $\theta=90^o$ but this a little different and is causing a little problem.
I approached this problem in the following matter and solved for $\theta= 90^o$ and got the wrong answer. Try and solve it for yourself.
$$\omega_h = \frac{2\pi}{12} ~~~~~ rad/hr $$
$$\omega_m = 2\pi ~~~~~ rad/hr $$
$$In ~time~ t,$$
$$\theta_h = \frac{2\pi}{12}t ~~~~~ rad $$
$$\theta_m = 2\pi t ~~~~~ rad $$
$$ |\theta_m - \theta_h|= \frac{\pi}{2} $$
$$\frac{2\pi t\times11}{12}=\frac{\pi}{2} $$
$$t=\frac{12}{2\times2\times11} $$
$$t= \frac{3}{11} $$
$ \therefore $ total number of such intervals in a day are, 
$$ \frac{24}{\frac{3}{11}} = 88 $$
But the answer is 44, So where has my maths gone wrong? One insight I have is that $\frac{3}{11}$ hours is time in which minute hand travels 90 + x and hour hand travels x degrees which is not what has been asked.
 A: Your equation $|\theta_m - \theta_h|= \frac{\pi}{2}$ does not measure the time interval between successive events, but the time between midnight and the first event after midnight. That is why the 'solution' $t=3/11$ seems to vary for other angles $\theta\neq\pi/2,$ and why it is incorrect even for $\theta=\pi/2.$
We do not need to measure the time between midnight and the first event, but the time between successive events. The fact that we do not care about the order of the two hands slightly complicates matters, so let us first treat the different problem where the oriented angle between the hands has to be equal to $\theta,$ i.e., the sign of $\theta$ matters.
Such events are evenly spaced in time, because the time interval between two such events is determined only by the relative angular velocity between the two hands. In fact, it is equal to $2\pi$ divided by their relative angular velocity $48\pi/\hbox{day}-4\pi/\hbox{day}=44\pi/\hbox{day}.$ The quotient is $1/22$th part of a day. That is the time for the hands to 'catch up with each other', i.e., to reach exactly the same relative oriented angle as in their initial position.
There are $22$ such events in a day. This is independent of the initial oriented angle chosen.
Now we pass to the original problem where the orientation of the angle does not matter. If you count only the absolute value of the angle $\theta,$ and $\theta$ is different from $0$ and from $\pm\pi,$ then the number of events doubles to $44,$ namely $22$ for every sign of $\theta.$
A: In 12 hours they make 11 overlappings, it is 1 and something 2 and something ... 11 and something. This is because the minute hand moves much faster than the hour hand and for $360^{\circ}+30^{\circ}$. In 12 hours the minute hand will cross $12 \cdot 360^{\circ} = 11 \cdot (360^{\circ} + 30^{\circ})$ so they overlap 11 times. The same goes for $180^{\circ}$ because for each overlap there is exactly one position within the same hour with $180^{\circ}$.
For any other angle, the minute hand and hour hand will position themselves twice within an hour so that they form that angle, one between the $n^{th}$ and $(n+1)^{th}$ overlap and another one between $(n-1)^{th}$ and $n^{th}$ overlap. It makes 22 positions.
In 24 h we have 44 such positions for angle that is not $0^{\circ}$ or $180^{\circ}$.
The problem in your equations is the distance you are trying to measure.
The first match is $\frac{\pi}{2}$, the next difference with the same relative position of hands is $2\pi+\frac{\pi}{2}$ and so on.
A: you checked only the first time, if you check that:
$k_1,k_2 \in \mathbb{Z}$
$|\theta_h - \theta_m| = \frac{\pi}{2}+2\pi k_1$ or $|\theta_h - \theta_m| = -\frac{\pi}{2}+2\pi k_2$
$\frac{22}{12} t = \frac{1}{2}+2k_1$ or $\frac{22}{12} t = -\frac{1}{2}+2k_1$
$0 \leq t\leq24$ because we check for 1 day.
$0 \leq \frac{22}{12} t \leq 44$
$0 \leq \frac{1}{2}+2k_1 \leq 44$ or $0 \leq -\frac{1}{2}+2k_2 \leq 44$
$0 \leq k_1 \leq 21$ or $1 \leq k_2 \leq 22$
$k_1$ and $k_2$ have 44 possible values.
A: $\frac 3{11}$ hours is the time it takes the minute hand to gain $90^\circ$ on the hour hand.  That will be the time from midnight (or noon) to the first $90^\circ$ point.  Having had one case you need the minute hand to gain $180^\circ$ to get to the next one, which takes $\frac 6{11}$ hours.
