Finding the exact time the hands of a clock form a given angle?

Question

Is there any method of finding what time a clock will form a certain angle? For example, what is the first time after 5:00 that the minute and hour hands of a clock will for a 54° angle? Thanks.

Answer 1

Let's think it out.

The minute hand travels $360 degrees/hour = 6 degrees/minute$.

The hour hand travels $360 degrees/12 hour = 30 degrees/hour = .5 degree/minute$

So the angle between the two hands is $5.5 degrees*minutes$.

It takes $m = 360/5.5= 65 \frac 5{11}$ minutes for the hour and minute to meet up again.

So the time it will take for the hands to be $\theta $ degrees appart will be $\theta/5.5$ minutes plus $k$($1$ hour $5 \frac 5{11}$ minutes).

So to find when the hands are $54$ degrees apart will be $54/5.5 = 9 \frac 9{11}$ minutes. So this will occur at $12:09\frac 9{11}$. $1:15 \frac 3{11}$. $2:20\frac 8{11}$. $3:26$. etc. If we need whole minutes. $3:26$ is the only acceptable answer.

If we need whole minutes, we can solve this by noting $\theta/5.5 = b + \frac k{11}$. If $k \le 5$ then we add $2k$($1$ hour $5 \frac 5{11}$ minutes) to get a $2k$ hours and $b + 11k$ minutes. If $k >5$ we add $(2k - 11)$($1$ hour $5 \frac 5{11}$ minutes) to get $2k - 12$ hours and $b + 11k $ minutes .

So for another example if we want to find to the minute when the difference of hands is $73$ degrees apart:

$73/5.5 = 146/11 = 14 \frac 3{11}= b + \frac k{11}$. So the time will be $2*k$ hrs and $b + 11k$ minutes or $2*3:14 + 11*3 = 6:47$.

Last thing we need to note is that we are not distinguish angles greater than $180$ degress. i.e. if the minute hand is 73 degrees past the hour hand is the same as the hour hand 73 degrees pas the minute hand. In that case we need to simple take the negative value.

So $6:47$ and $12:00 - 6:47 = 5:13$.

Answer 2

Full circle is 60 minutes and $360^\circ$ so $$ 54^\circ = 54^\circ \times \frac{60 \text{ min}}{360^\circ} = 9 \text{ min}. $$ You are starting with 5 pm, at 25 minutes distance, and every minute the minute hand moves $1$ tick closed but the hour hand moves $1/60$ tick away, so you gain $$ 1 - \frac{1}{60} = \frac{59}{60} \text{ ticks} $$ each minute. So $$ 25 - \frac{59x}{60} = 9, $$ can you solve the equation?

Answer 3

For a question of this type, "first time after $t_0$," there is a relatively straightforward procedure. Just ask and answer the following questions:

1. What is the angle between the hands at time $t_0$?
2. How much must the angle change to get to the desired angle?
3. How fast is the angle changing? (This one always has the same answer as long as the clock is a conventional 12-hour clock with minute and hour hands.)

From the answers to questions 2 and 3, you figure out how long it takes for the hands to reach the desired configuration, and add that to $t_0$ to get the time when the configuration occurs.

Just be careful that you correctly account for whether the minute hand has to catch up to the hour hand first and then get "ahead" by the given angle, or whether the minute hand can make that angle with the hour hand before catching up with it.

Answer 4

The equation for the angle between the hands: