The angle between two clock pointers Fun with math time
The other day a friend of mine asked me for this:

What is the value of the angle between two clock pointer when it's 11:50?

Of course the correct answer is not $60$ degrees, and it's quite clear why it's not.
Indeed, aside the motion of the minutes pointer there is also the motion of the hours pointer.
I calculated it, and I found out it's $55$ degrees. But I wasn't satisfied, so I decided to try to find a general formula to measure the angle between the two clock pointers at any hour.
My reasoning started with the "zero hour", which is 12:00 am (or pm, it's the same) in which the angle is $0$ degrees. Assuming the hours pointer is locked, $1$ hour (i.e. $60$ minutes) means a complete circle for the minutes pinter, namely $360°$. This means automatically that 
$$1\ \text{minute} = 6°$$
But the hours pointer is not locked, it moves! And quite elementarily we have that every $60$ minutes, the hours pointer moves about $1$ hour which in terms of steps is $\frac{1}{12}$ of a round angle.
$$1\ \text{hour} = 30°$$
So I calculated that from the total round angle of $2\pi$ or $360°$ we have to subtract a quantity equals to
$$\left(6M - \frac{M}{2}\right)$$
Where $M$ is the number of minutes; $6M$ measures the angle after $M$ minutes, and $\frac{M}{2}$ is due to the hours pointer motion. Indeed if every $60$ minutes the hours pointer moves about $30$ degrees, it means that every minutes, the hours pointer moves about $\frac{1}{2}$ degree. So after $M$ minutes it moves about $\frac{M}{2}$ degrees.
Up to now we have
$$\theta = \left[360 - \left(6M - \frac{M}{2}\right)\right]$$
that is
$$\theta = \left[360 - \frac{11}{2} M\right]$$
This formula is however incomplete because we have to take into account also the (I don't know how to call this) "reference starting point": this formula holds only from 12:00 to 1:00. But if we try to calculate it when it's 4:20 we would find
$$\theta = 250°$$
which is obviously wrong.
This is why we have to add one more term and one more condition: the term to add is 
$$30 H$$
where $H$ is the number of hours, in which we have to follow this convention:
$H = 0$ when it's 12:xy
$H = 1$ when it's 1:xy
$H = 2$ when it's 2:xy
and so on until $H = 11$$
So we came up with
$$\theta = \left(360 - \frac{11}{2}M\right) + 30H$$
But it's not over, since that result may exceed the value of $360$. When this happens, we need to subtract $360$ to get the right angle, so in the end:
$$\boxed{\theta = \left[\left(360 - \frac{11}{2}M\right) + 30H\right]_{360}}$$
In which the subscript "$360$" means indeed if the final value inside the square bracket is greater than $360$, then subtract $360$
Applying this to the example before, 4:20 we get:
$$M = 20 ~~~~~~~ H = 4$$
$$\theta = \left(360 - \frac{11\cdot 20}{2}\right) + 30\cdot 4 = 370$$
Since it's greater than $360$ we have finally
$$\theta = 370 - 360 = 10°$$
Now the questions
1) Is that a good formula, or we can improve it?
2) Is that a good formula or we can simplify it in a more "cute" form?
Thank you for all your time!
 A: If we are investigating the angle at a time in minutes, then we need to measure the hand poisiton at the minute for the hour hand and minute hand.
Obviously, the minute hand moves 6 degrees every minute, as you said.
Let us say the time is displayed as A:B where A is the hour, B is the minutes and this is strictly in 12 hour format.
Then the degrees (in the clockwise direction) for the 12 on the clock will be $6B^\circ$.
Now we come to the Hour hand. We will be investigating the position of the hour hand, every minute. Obviously, the hand moves $\frac{1}{12}360 = 30^\circ$ every hour. Therefore, in 1 minute, it should move $\frac{30}{60}^\circ = 0.5^\circ$ every minute.
Therefore, if A is the number of hours, then 60A is the number of minutes. Therefore at A hours, the hand would be at $$60A(0.5) = 30A^\circ$$
Now we need to take into account the additional number of minutes, i.e. B.
So B is going to be in  minutes, which means that $0.5B^\circ$ should be added to the $30A^\circ$
So we have that the hour hand is at $$(30A + 0.5B) ^\circ$$
So, now we know the angle for a time A:B.
So, the difference in the angle would be the absolute value of the difference of these = $$abs(30A + 0.5B - 6B) = |30A -5.5B|^\circ$$
We can test this out with the example you gave $4:20$
So the Hour hand would be at $$30(4) + 0.5(20) = 120 + 10 = 130^\circ$$
The minute hand would be at $$6(20) = 120^\circ$$
And the difference is $10^\circ$.
So the final solution is $$\theta = |30A - 5.5B|^\circ$$ where the time is written in the format $A:B$ and $A \in [0,11]$, $B \in [0,59]$
A: consider the clock as 0deg at 0hr and 360 degrees at 12 hours
and each hour occupies 30 degrees
Then the 11:50 angle is:
1.big hand=10/12*360=300degrees
MINUS
2.little hand=11 oclock posn + 50/60*(of the 30degrees of the 11th hour)
=11/12*360 + 5/6*30
=330 + 25
=355 degrees
THUS the differnce between the hands is 2.big hand- 1.little hand =355-300
                          =55 degrees

