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Can someone provide a solution for this question ...

Given the hours , minutes and seconds hands calculate the number of right angles the three hands make pairwise with respect to each other during a day... So it asks for the second and hour angle , minute and hour and second and minute

Thanks a lot..

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What do you mean by the number of right angles the three hands make? If you have three hands, you actually have 6 angles there, which come in pairs $\theta$ and $2\pi- \theta$. Are you asking for one of those angles to be $\frac{\pi}{2}$ or two of them? –  N. S. Oct 30 '11 at 21:01
And here is a hint, no matter what the problem is. If the hour hand is between $n$ and $n+1$, and the minute is perpendicular on it, how many possibilities are there? Minutes compared with seconds is also tricky, and hours compared with seconds is a little more subtle, but still easy. –  N. S. Oct 30 '11 at 21:03
An example would help clarify the kind of situation are you are wanting to enumerate. –  Austin Mohr Oct 30 '11 at 21:34
Yes I am ... Actually this was a question asked in an interview by Microsoft .. so i believe they wanted pairwise enumeration... For example there are 2 right angles each hour between the minute and hour hands... If we start at 12 one will occur sometime after 12:15 since the hour hand will also move... I am trying to find the general equation for the number ... –  user8250 Oct 30 '11 at 21:55

2 Answers 2

up vote 5 down vote accepted

Take two hands: a fast hand that completes $x$ revolutions per day, and a slow hand that completes $y$ revolutions per day. Now rotate the clock backwards, at a rate of $y$ revolutions per day: the slow hand comes to a standstill, and the fast hand slows down to $x-y$ revolutions per day. So the number of times that the hands are at right angles is $2(x-y)$.

The three hands make 2, 24, and 1440 revolutions per day, so the total is: $$2\times(24-2) + 2\times(1440-2) + 2\times(1440-24) = 5752$$

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Nice.${}{}{}{}$ –  Gerry Myerson Oct 31 '11 at 1:38

UPDATE: Okay, I'll revise my count and way of thinking:

Rather than considering things per-hour, I'll consider the number of times a hand "laps" another hand, that is to say the number of times it does 360* more than the other hand. For that to happen, the faster hand must have been in right angles twice. We know which hand is faster, so recompute:

Hour-Minute: 24 hours = 2 loops (hour hand), 24 hours = 24 loops (minute-hand), thus the minute hand passes the hour hand 22 times => 44 loops.

Hour-Second: 24 hours = 1440 loops (second-hand), thus the second hand passes the hour hand 1440 - 2 = 1438 times => 2876 loops.

Minute-Second: 24 loops (minute-hand) vs. 1440 loops (second-hand), thus we have 1416 passes => 2832 loops.

The total is 5752 loops.

Old answer:

We have 3 hands and therefore 3 "pairs" to consider. They are the hour-minute, hour-second, and minute-second.

Each hour, the minute hand will be at a right angle twice, so that's 48 right angles -- 2 for each hour of the day.

Each hour, the second hand will go around 60 times and have 2 right angles for each, that's 120 right angles per hour or 120 * 24 = 2880.

Each minute, the second hand will make 2 right angles with the hour hand, that's another 120 * 24 = 2880.

The total is 5808.

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I think you'll find that between 2 o'clock and 4 o'clock the hour and minute hands are at right angles only 3 times, not 4, so you're going to have to revise your count a bit. –  Gerry Myerson Oct 31 '11 at 0:37

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