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

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.

• There are no arc lengths here, just angles. At six o'clock the minute hand is at $0^{\circ}$ and the hour hand is at $180^{\circ}.$ The minute hand advances at $6^{\circ}$ per minute and the hour hand at $0.5^{\circ}$ per minute. No trigonometry or geometry involved, really. – saulspatz Feb 27 '18 at 21:02
• @g.kov Can you please use your expertise in drawing a sketch for this problem please?. That would be very appreciated!. :) – Chris Steinbeck Bell Feb 28 '18 at 16:17

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: $$5.5m - 180 = m \\ 4.5m = 180 \\ m = 40.$$

• 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.

• Can you make a drawing or a sketch so I can understand you better?. I made this request as part of the proposed answer. – Chris Steinbeck Bell Feb 28 '18 at 16:13
• How do I know the hour hand of the clock advances $0.5^{\circ}$ per minute?. Can you include this as part of your answer please?. – Chris Steinbeck Bell Feb 28 '18 at 17:04
• The first expression I get from the problem as states the angle between the minute hand and the hour hand. But the second is not very clear. We differentiate so the total number of minutes that would be transformed into angles is $6m-0.5m=5.5m$ Am i understanding that correctly?. Then the 180 degrees minus that angle would be the same as the angle between the hour hand and the minute hand makes?. – Chris Steinbeck Bell Feb 28 '18 at 17:23
• What the $5.5m$ tells you is that the minute hand gains $5.5^{\circ}$ on the hour hand for each minute that passes. The minute hand starts $180^{\circ}$ behind the hour hand, so that's why it's subtracted. When $5.5m-180=0$, the two hands are on top of one another (zero degrees difference). When the difference equals $m$, then the minute hand is ahead by $m$ degrees. – John Feb 28 '18 at 20:46