How many times to the hands of a clock form a 60 degree angle between noon and midnight on the same day?
Firstly im not sure weather they require the second hand to be included.
And secondly (excuse the pun) i have no clue how to start the problem.
Thanks.
On
Think of the minute hand and hour hand as running around a circular racetrack. The minute hand repeatedly passes the hour hand and loops around behind it. This happens 11 times from noon to midnight. So the question becomes: how many times is the distance along the track between the two hands 1/6 of the length of the track?
For each time the minute hand "laps" the hour hand, there were two times in between where the angle between the hands was 60°. The minute hand laps the hour hand 11 times from noon to midnight, so the hands form a 60° angle 22 times during that period.
On
Note that the minute hand starts in the same position as the hour hand. It catches up again a little after 1:00pm, between 1:00 and 2:00 (first); between 2:00 and 3:00 (second); between 3:00 and 4:00 (third); between 4:00 and 5:00 (fourth) ... between 10:00 and 11:00 (tenth) then catches up at 12:00 midnight (eleventh).
Between each pair of crossings, there are two occasions on which the angle is $60^{\circ}$.
Note that the $11$ crossings occur over $12$ hours and are hence $\cfrac {12}{11}$ hours apart - or one hour and $\cfrac {60}{11}$ minutes apart. The first $60^\circ$ crossing occurs one sixth of the time between crossings, $\cfrac 16\cdot \cfrac{12}{11}=\cfrac 2{11}$ hours after each crossing. The second is after five sixths of the time between crossings or $\cfrac {10}{11}$ hours since the last crossing.
Hint
Total angle swept by the pins on a clock $ = 360^{\ \text{o}} $
Smallest unit on clock (for recreational questions, second hand is NOT taken into consideration) $ = 1 $ minute
$ \therefore $ Area swept per minute $ \implies 6^{\text{ o}} $.
Hence, to generate an angle of $ 60^{\text{ o}} $ (between hour and minute hands), they should be $ 10 $ units apart.