How can I relate the lengths in an analog watch and the angles in sexagesimal system?

Question

The problem is as follows:

In an analog watch; at what time immediately after 6 am, the number of minutes elapsed after 6 am is equal to the number of sexagesimal degrees the minute hand of the watch is ahead of the hour hand?

I'm puzzled on this situation as I don't know how can I relate the length of the arc and the minutes the minute hand advances. What I found were these relationships.

$1\,\textrm{min}=6 ^{\circ}$

The minutes the hour hand advances equates the following:

$12\,\textrm{min}=30 ^{\circ}$

I'm not sure if to solve this problem is required geometry or trigonometry, but since there is some connection between the arc lengths I'd like if somebody could help me with a drawing or diagram to clearly understand this situation better.

Answer 1

The angle of the hour hand (in degrees) is $180+0.5m$, where $m$ is the number of minutes elapsed. The angle of the minute hand is $6m$, so the number of minutes the minute hand is ahead of the hour hand is $5.5m - 180$.

Can you take it from here?

Spoiler:

Setting this equal to $m$ should give you your answer: $$5.5m - 180 = m \\ 4.5m = 180 \\ m = 40.$$

Regarding your questions in the comments, here are some hints:

• Can you make a drawing or a sketch so I can understand you better?

Draw an analog clock that shows six o'clock. Measure the angle clockwise starting at the 12. The minute hand is at $0^{\circ}$ and the hour hand is at $180^{\circ}$

Now draw a clock that shows $6:05$. Don't worry about exactly where the hour hand is. The minute hand, now pointing at the 1, makes an angle of $360^{\circ}/12 = 30^{\circ}.$

• How do I know the hour hand of the clock advances $0.5^{\circ}$ per minute?

From above, we see that going from the 12 to the 1 is $30^{\circ}.$ The hour hand goes through this angle in one hour. So divide by $60$ to get how far the hour hand goes in a minute.