A problem regarding the hands of a clock Q) Between 12 noon and 1 PM, there are two instants when the hour hand and the
minute hand of a clock are at right angles. The difference in minutes between
these two instants is what?
I am unable to find that certain angle that'll give me the difference in minutes, I have tried drawing figures and finding the angles but it seems like it can vary and still add up to the same conditions as mentioned...
 A: Note that the minute hand and the hour hand are both moving with a constant angular velocity. At $12\text{pm}$, the angle between them is zero, and then after some amount of time $t$, the angle between them is $90^{\circ}$. It follows that at $2t$ the angle between them will be $180^{\circ}$; at $3t$ the angle between them will be $270^{\circ}$; and at $4t$ the minute hand and hour hand will meet again. We want to find $t$ and $3t$.
Now note that the minute hand and the hour hand meet once every hour between $1\text{pm}$ and $11\text{pm}$, and at $12\text{am}$, the minute hand and hour hand meet again, for the $11^{th}$ time. It follows that
$$11\cdot4t=12\text{ hours.}$$
Hence:
$$\begin{align*}
t &= \frac{3}{11}\text{ hours}=16+\frac{4}{11}\text{ minutes} \\
\\
3t &= \frac{9}{11}\text{ hours}=49+\frac{1}{11}\text{ minutes}
\end{align*}$$
So the first time is between $12$:$16\text{pm}$ and $12$:$17\text{pm}$, and the second time is between $12$:$49\text{pm}$ and $12$:$50\text{pm}$. The difference between the two times is
$$3t-t=2t=\frac{6}{11}\text{ hours}=32+\frac{8}{11}\text{ minutes}.$$
A: 32 minutes 44 seconds

Here's a simplified (i.e., probably not the best) way of looking at it assuming continuous, constant movement of the hands of the clock. One nice, simplifying thing is that the hour hand and minute hand start at the same point, let's say $(0,0)$.
Let's use $(t,\theta)$ to denote the internal angle of the circle $\theta$ swept out over time $t$, where $t$ is measured in minutes. For the hour hand, we start at $(0,0)$ and end at $(60,\frac{1}{12}\cdot2\pi)$ over the course of an hour. We can let the function $\theta_h(t)=\frac{\pi t}{360}$ describe such motion. Similarly, for the minute hand, we start at $(0,0)$ and end at $(60,2\pi)$ over the course of an hour. We can let the function $\theta_m(t)=\frac{\pi t}{30}$ model this motion. 
Hence, we are essentially looking for a $t$ for which $\theta_m-\theta_h=\frac{\pi}{2}$ as well as a $t$ for which $\theta_m-\theta_h=\frac{3\pi}{2}$. Or, more helpfully, we are trying to find a $t$ for which
$$
\frac{\pi t}{30}-\frac{\pi t}{360}=\frac{\pi}{2},\frac{3\pi}{2},
$$
and doing some basic algebra shows that what we get are solutions when $t=\frac{180}{11}, \frac{540}{11}$; hence, the difference in minutes between these two instances is 
$$
\frac{540}{11}-\frac{180}{11}=\frac{360}{11}=32\frac{8}{11}\;\text{minutes}
$$
or, more simply, 32 minutes and 44 seconds.
Side note: The hour and minute hands will be at right angles at (roughly) 12:16:22pm and 12:49:05pm.

As said above, this is probably not the cleanest way of approaching it, but maybe it will be somewhat clearer.
A: The easy way:
You don't actually need to find the two times when the hour and minute hand are at right angles.  Just note that the difference between these two times is how long it takes for the minute hand to move $180^\circ$ farther than the hour hand has moved.  The minute hand moves at $6^\circ$ per minute, and the hour hand moves at $\frac{360}{12\cdot 60}^\circ=\frac{1}{2}^\circ$ per minute.  So the time required for the minute hand to advance $180^\circ$ farther than the hour hand is $\frac{180}{6-1/2}=\frac{360}{11}$ minutes. 
A: At $n$ minutes after 12 noon, the minute-hand is at an angle $\theta_m$ with the vertical direction, given by the ratio 
$$\frac{\theta_m}{360}=\frac n{60} \implies \theta_m = 6n$$
and the hour-hand is at an angle $\theta_h$ with the vertical direction
$$\frac{\theta_h}{\frac{360}{12}}=\frac {n}{60} \implies \theta_h = \frac12{n}$$
For the two hands at a right angle, we have
$$\theta_m-\theta_h = 6n - \frac12 n=90, \>270$$
which yields $n= \frac{180}{11},\>\frac{540}{11}$. Thus, the difference in minutes is
$$ \frac{540}{11}-\frac{180}{11}=\frac{360}{11}=32.6 \>\text{mins}$$
A: Deep breath.  
Let's divide the clock into $60$ minute marks.  $90^0$ is one quarter of a circle so it is $15$ of those minute marks.  So if the minute hand is at the $m$ mark the hour hand is at the $m \pm 15$ mark.
But we should consider that if $m> 45$ then $m + 15 > 60$ and we actually have gone a full hour and returned to the beginning.  So if the minute hand is at the $m$ mark and $m > 45$ then the hour hand will be at the $m + 15 -60 = m -45$ mark.  And if $m < 45$ then the hour hand will be at the $m \pm 15$ mark.
But we are only taking time between $12$ and $1$ so the hour hand will only have travelled $\frac 1{12}$ of the circle of $5$ marks.  So the hour hand at the $m + 15$ mark is impossible.
So the hour hand will either be at the $m - 15$ mark (and it's roughly 15 minutes after 12) or it will be ant the $m-45$ (and it's roughtly 45 minutes after 12).
.......
Okay, let's start
.......
Okay.  After $x$ minutes the minute hand will be $x$ mark.  
And as the minute hand takes $1$ hour to go a full circle and the hour hand takes $12$ hours, the hour hand travels only $\frac 1{12}$ as fast.  SO
After $x$ minutes the hour hand will be at the $\frac x{12}$ mark$.
So we have either 
1) $x -15 = \frac x{12}$ or
2) $x -45 = \frac x{12}$ 
If 1)
$x-15 = \frac x{12}$
$12x - 180 = x$
$11x = 180$
$x = \frac{180}{11}= 16\frac 4{11}$
So this will occur at $16\frac 4{11}$ minutes after $12$.  Or as there are $60$ seconds in a minute and $12: 16: 21\frac 9{11}$.
2) Same thing
$11x =540$ and $x = 49 \frac 1{11}$ so this will occur at $49\frac 1{11}$ minutes after $12$ of $10\frac {10}{11}$ minutes before $1$ or and $12:49:05\frac 5{11}$.
