Tickity-Tockity

If the clock hands meet at midnight, two hours and twenty minutes ago was $9:40\text{ PM}$ . Therefore, the minute hand is on the $8$ and the hour hand is $\frac{2}{3}$ of the way between $9$ and $10$ :

The twelve markings divide the clock into $\frac{360^\circ}{12} = 30^\circ$ intervals, so the answer is $30^\circ+20^\circ=\boxed{50^\circ}$ .

In a 12-hour period, the hour hand travels one revolution and the minute hand travels 12 revolutions. The difference $\Delta \phi$ between the hands travels therefore 11 full revolutions. Since this angle increases at a constant rate, the rate at which $\Delta \phi$ increases is $\frac {d(\Delta \phi)}{dt} = \frac{11 \times 360^\circ}{12\ \text{hours}} = 330^\circ/\text{hour}.$ In the time of 2 hours, 20 minutes, the difference is $\Delta \phi = 330^\circ/\text{hour} \times 2\tfrac13\ \text{hours} = 770^\circ = 2\times 360^\circ + 50^\circ.$ Thus the hands will be $\boxed{50^\circ}$ apart. (This is true both for 2:20 pm and 9:40 am.)

The hour hand travels at a rate of $\frac{360°}{12} = 30°$ clockwise every hour. Therefore, $2$ hours and $20$ minutes ago, the hour hand was $2 \frac{1}{3} \cdot 30° = 70°$ counterclockwise from its current position.

The minute hand travels at a rate of $\frac{360°}{60} = 6°$ clockwise every minute. Therefore, $2$ hours and $20$ minutes ago, the minute hand was $2$ full revolutions and $20 \cdot 6° = 120°$ counterclockwise from its current position.

The angle between the two hands is then the difference of $120° - 70° = \boxed{50°}$ .

First, since the clock can be rotated, we can assume the hour and minute hand meet at the "top", that is, 12:00.

Furthermore, since the clock is symmetrical, 2 hours and 20 minutes behind will result in the same angle as 2 hours and 20 minutes ahead, so we can solve this by finding the angle between the clock hands at 2:20.

At 2:20, the minute hand is at the 4. The hour hand is at the 2 plus an additional third of the way to the 3 (this is because 20 minutes is a third of an hour).

The 360 degrees of a clock are divided by the face numbers into 12 equal parts, so there are $360/12 = 30$ degrees between face numbers. The distance between the hour hand and the 3 is two-thirds of that, that is, $30 \left(\frac{2}{3}\right) = 20$ degrees. So we have a total angle of $20 + 30 = 50$ degrees between the hour and minute hands.

Here is a solution from let's say a more physical perspective.

Since it is easier for me to look forwards in time rather than backwards, the problem doesn't change if we ask ourselves what will be the angle between the two hands 2 hours and 20 minutes from now if they start from the same position. This is exactly the same as asking the question if the two hands are in the same position, what was the angle between them 2 hours and 20 minutes ago.

The clock can be abstracted away. We are not interested in hours, minutes, seconds or their position on the clock. This problem can simply be reduced to two hands which are rotating at different, but constant angular velocities that are starting from the same position.

The clock hand makes one full revolution (or 360 degrees) after 12 hours so the angular velocity of the hour hand is: $\omega_h = \frac{360^\circ}{12 \text{h}} = 30 \frac{\text{deg}}{\text{hour}}$

The minute hand makes one full revolution (or 360 degrees) after 60 minutes (or one hour) so the angular velocity of the minute hand is: $\omega_m = \frac{360^\circ}{1 \text{h}} = 360 \frac{\text{deg}}{\text{hour}}$

Time $t$ is 2 hours and 20 minutes, but let's write this in hours. 20 minutes is 1/3 of a full hour, so: $t = 2\text{ h}+1/3 \text{ h} = \frac{7}{3} \text{ hours}$

After 7/3 hours, the hour hand will move: $\theta_h = \omega_h \times t = 30\times \frac{7}{3} = 70 \text{ deg}$

and the minute hand will move: $\theta_m = \omega_m \times t = 360\times \frac{7}{3} = 840 \text{ deg} = 360 + 360 + 120 \text{ deg} = 120 \text{ deg}$

Considering angles larger than 360 degrees doesn't make any sense in this case because after 2 hours and 20 minutes the minute hand will make two complete revolutions and land 120 degrees from the starting position.

Finally, the angle difference between the two hands is equal to the starting difference (which is 0 in this case) and the difference between the two angles we just calculated:

$\Delta \theta (t) = \Delta \theta_0 + \theta_m(t) - \theta_h(t) = 0 + 120 - 70 = 50 \text{ deg}$

Now we have a formula for any arbitrary starting position and any arbitrary time: $\Delta \theta (t) = (\Delta \theta_0 + \theta_m(t) - \theta_h(t)) \text{ mod } 360 = (\Delta \theta_0 + (\omega_m-\omega_h)\times t)\text{ mod } 360 = (\Delta \theta_0 + 330 \times t)\text{ mod } 360$

For example: if the clock shows 4 hours and 15 minutes, what will be the angle difference after 3 hours and 40 minutes?

$\Delta \theta_0 = -\theta_h(0) + \theta_m(0) = -4.25\times30 + \frac{15}{5}\times 30 = -37.5 \text{deg}$ $t = \frac{11}{3} \text{h}$ $\Delta \theta (t) = (\Delta \theta_0 + \theta_m(t) - \theta_h(t)) \text{ mod } 360 = (-37.5 + 330\times\frac{11}{3})\text{ mod } 360 = 1172.5\text{ mod } 360=92.5 \text{deg}$

In 1 Hour, the Hour hand rotates by $30^\circ$ .Therefore, in 1 minute it rotates $\frac{30^\circ}{60}$ = $\frac{1}{2}^\circ$ .

Similarly, the minute hand in 1 minute rotates by $\frac{360^\circ}{60}$ = $6^\circ$ .

The difference between the speed of rotation of hour and minute hand is $6^\circ - \frac{1^\circ}{2} = 5\frac{1^\circ}{2}$ .

What the above statements imply is that the minute hand covers $\frac{11^\circ}{2}$ more than the hour hand per minute.

Now, 2 Hours and 20 Minutes = 140 Minutes and both the hands are at the same position initially.

Therefore,the difference between the hour hand and the minute hand in 140 minutes will be $140 * \frac{11^\circ}{2} = 770^\circ$ .

Since $360^\circ$ will mean a complete circle , we can clearly remove 2 circles from $770^\circ$ .

Giving us the answer $770^\circ - 720^\circ = \boxed{50^\circ}$ .

Let's assume that the minute and hour hands have met up at exactly noon (12:00 p.m.). Two hours and twenty minutes ago, the time was 9:40 a.m.

Let us calculate how many degrees the minute hand has rotated about the center of the clock face since 9:00. Every number marking along the circumference corresponds to a 30º rotation clockwise and is situated 5 minutes apart from the next marking. After 40 minutes, the minute hand will have traveled 8 of these divisions around, making an angle of 8(30º) = 240º.

At 9:00, the hour hand had traveled 9(30º) = 270º around since midnight, but how far does it rotate after 40 minutes? Well, the hour hand travels 30º for every 360º that the minute hand moves around. Since 40 minutes is $\frac{2}{3}$ of 60 minutes, and 240º is $\frac{2}{3}$ of 360º, we can deduce that the minute hand has traveled $\frac{2}{3}$ of the way around the circumference of the clock face ( $\frac{2}{3}$ of 360º). Since the rotation rates of the two minute hands are proportional to each other (30º per 360º), we can correctly say that the hour hand will travel $\frac{2}{3}$ of 30º in 40 minutes, which is 20º. Adding on the extra 270º from its rotation since midnight, we find that the hour hand has traveled 290º around the clock face.

Thus the angle between the two hands is 290º – 240º = 50º.

Note that every minute, the angle between the hour and minute hand increases by $5.5^\circ$ . So, the angle between them is $(5.5 \cdot (2 \cdot 60+20))^\circ=770^\circ$ . This is also equivalent to $770^\circ-720^\circ=\boxed{50^\circ}$

Imagine both hands start on the twelfth hour. Move the hour hand back 2 hours, and move the minute hand back 20 minutes. The amount of time between the hour hand and the minute hand will be 1 hour and 40 minutes, or 1.66... hrs. Now, find the ratio of the amount of time separating the two hands to the total hours on the clock.

1.66... hrs / 12 hrs = 0.1388...

Now, multiply your ratio by 360 degrees (the clock is a circle, so it encloses 360 degrees) to determine the angle between the two hands.

360 degrees * 0.1388... = 50 degrees.

I am so sorry. This isn't my problem, someone at brilliant or a hacker changed the problem and answers, even though it is still showing as mine, and has the same title. Does anyone know what could have happened?

If the hour and minute hands are exactly on top of each other, then they are both pointing at 12. 2 hours and 20 minites ago it was 9:40. So, the hour hands is between 9 and 10, 2/3 of the way to 10, and the minute hand is at 8. Divide each hour into 3 divisions, so the number of divisions between the two hands is 5 (3 divisions from 8-where the minute hand is standing- to 9, and 2 diviosns from 9 to where the hour hand is standing). Number of total divions is 3x12=36 So the angle is (5/36)x360=50 degrees.