To find angular distance between the minute and hour hand, first find the position of each. Using $\displaystyle 12\textup{:}00$ as a reference (both hands straight up), we can calculate the difference in degree more easily.

First, remember that a circle contains $\displaystyle 360^{\circ}$, and therefore for the minute hand, each minute past the hour takes up $\displaystyle \frac{360^{\circ}}{60} = 6^{\circ}$ of angular distance.

For the hour hand, each hour in a 12-hour cycle takes up $\displaystyle \frac{360^{\circ}}{12} = 30^{\circ}$ of angular distance, which means each minute takes up $\displaystyle \frac{30^{\circ}}{60} = (\frac{1}{2})^{\circ}$ of distance for the hour hand.

Thus, our distance from $\displaystyle 12:00$ (our reference angle) in degrees for the minute hand can be expressed as $\displaystyle d(M) = (6m)^{\circ}$, where $\displaystyle m$ is the number of minutes that have passed.

Likewise, our distance from $\displaystyle 12:00$ in degrees for the hour hand can be expressed as $\displaystyle d(H) = (\frac{m}{2})^{\circ}$, where $\displaystyle m$ is again the number of minutes that have passed.

Then, all we need to do is find the positive difference between these two measurements, $\displaystyle \left | d(M) -d(H) \right |$, and we have our angle.

This looks more complicated than it is! Fortunately, once our equation is set up, the problem is as easy as plug and play. First, let's find the measure of the minor arc between the minute hand and the reference. We can ignore hours (as each hour returns the minute hand to our reference angle) and just look at minutes.

$\displaystyle d(M) = (6m)^{\circ} = (6\cdot39)^{\circ} = 234^{\circ}$

Now, find the same distance, but for the hour hand. Here, we must consider both hours and minutes by using a fraction:

$\displaystyle d(H) = (\frac{m}{2})^{\circ} = (\frac{(11\cdot60)+39}{2})^{\circ} = (\frac{699}{2})^{\circ}$

Lastly, we find the difference between these two references (remembering that our answer should be positive):

$\displaystyle \left | d(M) -d(H) \right | = \left \| 234^{\circ} - (\frac{699}{2})^{\circ} \right \| = (\frac{231}{2})^{\circ}$

Thus, the hands are $\displaystyle (\frac{231}{2})^{\circ}$ apart at $\displaystyle 11:39$.

To find angular distance between the minute and hour hand, first find the position of each. Using $\displaystyle 12\textup{:}00$ as a reference (both hands straight up), we can calculate the difference in degree more easily.

First, remember that a circle contains $\displaystyle 360^{\circ}$, and therefore for the minute hand, each minute past the hour takes up $\displaystyle \frac{360^{\circ}}{60} = 6^{\circ}$ of angular distance.

For the hour hand, each hour in a 12-hour cycle takes up $\displaystyle \frac{360^{\circ}}{12} = 30^{\circ}$ of angular distance, which means each minute takes up $\displaystyle \frac{30^{\circ}}{60} = (\frac{1}{2})^{\circ}$ of distance for the hour hand.

Thus, our distance from $\displaystyle 12:00$ (our reference angle) in degrees for the minute hand can be expressed as $\displaystyle d(M) = (6m)^{\circ}$, where $\displaystyle m$ is the number of minutes that have passed.

Likewise, our distance from $\displaystyle 12:00$ in degrees for the hour hand can be expressed as $\displaystyle d(H) = (\frac{m}{2})^{\circ}$, where $\displaystyle m$ is again the number of minutes that have passed.

Then, all we need to do is find the positive difference between these two measurements, $\displaystyle \left | d(M) -d(H) \right |$, and we have our angle.

This looks more complicated than it is! Fortunately, once our equation is set up, the problem is as easy as plug and play. First, let's find the measure of the minor arc between the minute hand and the reference. We can ignore hours (as each hour returns the minute hand to our reference angle) and just look at minutes.

$\displaystyle d(M) = (6m)^{\circ} = (6\cdot40)^{\circ} = 240^{\circ}$

Now, find the same distance, but for the hour hand. Here, we must consider both hours and minutes by using a fraction:

$\displaystyle d(H) = (\frac{m}{2})^{\circ} = (\frac{(4\cdot60)+40}{2})^{\circ} = (\frac{280}{2})^{\circ} = 140^{\circ}$

Lastly, we find the difference between these two references (remembering that our answer should be positive):

$\displaystyle \left | d(M) -d(H) \right | = \left \| 240^{\circ} - 140^{\circ} \right \| = 100^{\circ}$

Thus, the hands are $\displaystyle 100^{\circ}$ apart at $\displaystyle 4:40$.

To find angular distance between the minute and hour hand, first find the position of each. Using $\displaystyle 12\textup{:}00$ as a reference (both hands straight up), we can calculate the difference in degree more easily.

First, remember that a circle contains $\displaystyle 360^{\circ}$, and therefore for the minute hand, each minute past the hour takes up $\displaystyle \frac{360^{\circ}}{60} = 6^{\circ}$ of angular distance.

For the hour hand, each hour in a 12-hour cycle takes up $\displaystyle \frac{360^{\circ}}{12} = 30^{\circ}$ of angular distance, which means each minute takes up $\displaystyle \frac{30^{\circ}}{60} = (\frac{1}{2})^{\circ}$ of distance for the hour hand.

Thus, our distance from $\displaystyle 12:00$ (our reference angle) in degrees for the minute hand can be expressed as $\displaystyle d(M) = (6m)^{\circ}$, where $\displaystyle m$ is the number of minutes that have passed.

Likewise, our distance from $\displaystyle 12:00$ in degrees for the hour hand can be expressed as $\displaystyle d(H) = (\frac{m}{2})^{\circ}$, where $\displaystyle m$ is again the number of minutes that have passed.

Then, all we need to do is find the positive difference between these two measurements, $\displaystyle \left | d(M) -d(H) \right |$, and we have our angle.

This looks more complicated than it is! Fortunately, once our equation is set up, the problem is as easy as plug and play. First, let's find the measure of the minor arc between the minute hand and the reference. We can ignore hours (as each hour returns the minute hand to our reference angle) and just look at minutes.

$\displaystyle d(M) = (6m)^{\circ} = (6\cdot2)^{\circ} = 12^{\circ}$

Now, find the same distance, but for the hour hand. Here, we must consider both hours and minutes by using a fraction:

$\displaystyle d(H) = (\frac{m}{2})^{\circ} = (\frac{(6\cdot60)+2}{2})^{\circ} = (\frac{362}{2})^{\circ} = 181^{\circ}$

Lastly, we find the difference between these two references (remembering that our answer should be positive):

$\displaystyle \left | d(M) -d(H) \right | = \left \| 181^{\circ} - 12^{\circ} \right \| = 169^{\circ}$

The problem asked for the major arc, so our answer is actually $\displaystyle 360-169= 191^{\circ}$

Thus, the hands are $\displaystyle 191^{\circ}$ apart at $\displaystyle 6\textup{:}02$.

To find angular distance between the minute and hour hand, first find the position of each. Using $\displaystyle 12:00$ as a reference (both hands straight up), we can calculate the difference in degree more easily.

First, remember that a circle contains $\displaystyle 360^{\circ}$, and therefore for the minute hand, each minute past the hour takes up $\displaystyle \frac{360^{\circ}}{60} = 6^{\circ}$ of angular distance.

For the hour hand, each hour in a 12-hour cycle takes up $\displaystyle \frac{360^{\circ}}{12} = 30^{\circ}$ of angular distance, which means each minute takes up $\displaystyle \frac{30^{\circ}}{60} = (\frac{1}{2})^{\circ}$ of distance for the hour hand.

Thus, our distance from $\displaystyle 12:00$ (our reference angle) in degrees for the minute hand can be expressed as $\displaystyle d(M) = (6m)^{\circ}$, where $\displaystyle m$ is the number of minutes that have passed.

Likewise, our distance from $\displaystyle 12:00$ in degrees for the hour hand can be expressed as $\displaystyle d(H) = (\frac{m}{2})^{\circ}$, where $\displaystyle m$ is again the number of minutes that have passed.

Then, all we need to do is find the positive difference between these two measurements, $\displaystyle \left | d(M) -d(H) \right |$, and we have our angle.

This looks more complicated than it is! Fortunately, once our equation is set up, the problem is as easy as plug and play. First, let's find the measure of the minor arc between the minute hand and the reference. We can ignore hours (as each hour returns the minute hand to our reference angle) and just look at minutes.

$\displaystyle d(M) = (6m)^{\circ} = (6\cdot5)^{\circ} = 30^{\circ}$

Now, find the same distance, but for the hour hand. Here, we must consider both hours and minutes by using a fraction:

$\displaystyle d(H) = (\frac{m}{2})^{\circ} = (\frac{(11\cdot60)+5}{2})^{\circ} = (\frac{665}{2})^{\circ} = (332 \frac{1}{2})^{\circ}$

Lastly, we find the difference between these two references (remembering that our answer should be positive):

$\displaystyle \left | d(M) -d(H) \right | = \left \| 30^{\circ} - (332\frac{1}{2})^{\circ} \right \| = (302\frac{1}{2})^{\circ}$

Thus, the hands are $\displaystyle (302\frac{1}{2})^{\circ}$ apart at $\displaystyle 11:05$.

To find angular distance between the minute and hour hand, first find the position of each. Using $\displaystyle 12:00$ as a reference (both hands straight up), we can calculate the difference in degree more easily.

First, remember that a circle contains $\displaystyle 360^{\circ}$, and therefore for the minute hand, each minute past the hour takes up $\displaystyle \frac{360^{\circ}}{60} = 6^{\circ}$ of angular distance.

For the hour hand, each hour in a 12-hour cycle takes up $\displaystyle \frac{360^{\circ}}{12} = 30^{\circ}$ of angular distance, which means each minute takes up $\displaystyle \frac{30^{\circ}}{60} = (\frac{1}{2})^{\circ}$ of distance for the hour hand.

Thus, our distance from $\displaystyle 12:00$ (our reference angle) in degrees for the minute hand can be expressed as $\displaystyle d(M) = (6m)^{\circ}$, where $\displaystyle m$ is the number of minutes that have passed.

Likewise, our distance from $\displaystyle 12:00$ in degrees for the hour hand can be expressed as $\displaystyle d(H) = (\frac{m}{2})^{\circ}$, where $\displaystyle m$ is again the number of minutes that have passed.

Then, all we need to do is find the positive difference between these two measurements, $\displaystyle \left | d(M) -d(H) \right |$, and we have our angle.

This looks more complicated than it is! Fortunately, once our equation is set up, the problem is as easy as plug and play. First, let's find the measure of the minor arc between the minute hand and the reference. We can ignore hours (as each hour returns the minute hand to our reference angle) and just look at minutes.

$\displaystyle d(M) = (6m)^{\circ} = (6\cdot14)^{\circ} = 84^{\circ}$

Now, find the same distance, but for the hour hand. Here, we must consider both hours and minutes by using a fraction:

$\displaystyle d(H) = (\frac{m}{2})^{\circ} = (\frac{(4\cdot60)+14}{2})^{\circ} = (\frac{254}{2})^{\circ} = 127^{\circ}$

Lastly, we find the difference between these two references (remembering that our answer should be positive):

$\displaystyle \left | d(M) -d(H) \right | = \left \| 84^{\circ} - 127^{\circ} \right \| = 43^{\circ}$

The problem asked for the major arc, so our answer is actually $\displaystyle 360-43= 317^{\circ}$

Thus, the hands are $\displaystyle 317^{\circ}$ apart at $\displaystyle 4:14$.

First, remember that a circle contains $\displaystyle 360^{\circ}$, and therefore for the minute hand, each minute past the hour takes up $\displaystyle \frac{360^{\circ}}{60} = 6^{\circ}$ of angular distance.

Thus, our total distance from our reference angle can be found as $\displaystyle d(m) = 6m^{\circ}$, where $\displaystyle m$ is the number of minutes that have elapsed.

In this case, we simply find the difference of the two times, remembering to first convert hours to minutes:

$\displaystyle 5:25-3:30 = 325 -210 = 115$

Now, plug our answer in minutes into our equation:

 $\displaystyle d(m) = 6m^{\circ} = 6(115)^{\circ} = 690^{\circ}$

So, our minute hand has moved $\displaystyle 690^{\circ}$ in total.

First, remember that a circle contains $\displaystyle 360^{\circ}$, and therefore for the minute hand, each minute past the hour takes up $\displaystyle \frac{360^{\circ}}{60} = 6^{\circ}$ of angular distance.

Thus, our total distance from our reference angle can be found as $\displaystyle d(m) = 6m^{\circ}$, where $\displaystyle m$ is the number of minutes that have elapsed.

In this case, we simply find the difference of the two times, remembering to first convert hours to minutes:

$\displaystyle 11\textup{:}19-5\textup{:}17 = 679 -317 = 362$

Now, plug our answer in minutes into our equation:

 $\displaystyle d(m) = 6m^{\circ} = 6(362)^{\circ} = 2172^{\circ}$

So, our minute hand has moved $\displaystyle 2172^{\circ}$ in total.

First, remember that an hour hand moves $\displaystyle \frac{360^{\circ}}{12} = 30^{\circ}$ for each hour that passes, and therefore for the hour hand, each minute past the hour takes up $\displaystyle \frac{30^{\circ}}{60} = (\frac{1}{2})^{\circ}$ of angular distance.

Thus, our total distance from our reference angle can be found as $\displaystyle d(h) = (\frac{m}{2})^{\circ}$, where $\displaystyle m$ is the number of minutes that have elapsed.

In this case, we simply find the difference of the two times, remembering to first convert hours to minutes:

$\displaystyle 8\textup{:}38-4\textup{:}10 = 518 -250 = 268$

Now, plug our answer in minutes into our equation:

 $\displaystyle d(h) = (\frac{m}{2})^{\circ} = (\frac{268}{2})^{\circ} = 134^{\circ}$

So, our hour hand has moved $\displaystyle 134^{\circ}$ in total.

When the clock reads $\displaystyle 3:00$, the hour hand is on the $\displaystyle 3$, and the minute hand is on the $\displaystyle 12.$  If we think about this as a fraction, there are twelve spots the hour hand can be on, which we means we are on the $\displaystyle \frac{3}{12}$ position.  

Since there are $\displaystyle 360 \; degrees$ in a circle, the angle can simply be found by multiplying this fraction by the number of degrees in a circle:

$\displaystyle \frac{3}{12}*360\;degrees=90\;degrees$

Alternatively, if the clock reads $\displaystyle 3:00$, the angle the clock reads is visually $\displaystyle \frac{1}{4}$ of the entire clock, which has $\displaystyle 360\;degrees$.

  $\displaystyle \frac{1}{4}*360\;degrees=90\;degrees$.

To find the degrees of a clock hand, first find the angle between each hour-long sections. Since there are $\displaystyle 12$ evenly spaced sections, we find that each section has an angle of: $\displaystyle 360$^{\circ}$/12=30$^{\circ}$$. at $\displaystyle \textup{11:20}$ the hour hand has gone one-third of the way between the $\displaystyle 11$and $\displaystyle 12$. Thus there are two-thirds of $\displaystyle 30^{\circ}$ between the hour hand and the $\displaystyle 12$. $\displaystyle \left (\frac{2}{3} \right )*30^{\circ} = 20^{\circ}$.

There are $\displaystyle 120^{\circ}$ between $\displaystyle 12$ and $\displaystyle 4$, where the minute hands is. Thus there's a total of $\displaystyle 20^{\circ} + 120^{\circ} = 140^{\circ}$ between the hands.