probability

Question

Find all exact times on a clock when the hands form a right angle. for example, 4:05 seems to be an answer, but it is NOT. more precisely, the time 4:05 and 5/11 minutes forms a right angle. 

Answer 1

The hour hand moves at a rate of 360(degrees)/12hrs or 30/hr which we'll write in minutes: 0.5/minute.

 

The minute hand moves at a rate of 360/hr or 6/minute.

 

We'll start with the 12 o'clock hour: You know that there are two points at which the minute and hour hand will be 90 degrees apart, the first time when the minute hand has just passed the hour hand, and the second time when the minute hand is coming up behind the hour hand again.

 

We can define two functions to represent the degrees away from the 12 each hand will be during this hour: 

 

For the hour hand, it will be 0.5t where t is in minutes. For the minute hand it will be 6t during the first half of the hour, and 360 - 6t during the second half of the hour.

 

To find how far apart the hands are each time, take the difference of these functions in the first half hour, and their sum in the second half hour, and set them equal to 90. This will tell you what t will be each time.

 

e.g.

 

First time:

 

6t - 0.5t = 5.5t = 90

 

So 11t = 180 

 

Then t = 16 and 4/11 of a minute. 

 

So the first time this happens in the noon hour is 12:16 and 4/11 of a minute.

 

Second time:

 

360 - 6t + + 0.5t = 360 - 5.5t = 90

 

Then 5.5t = 270

 

So 11t = 540

 

Then t = 49 and 1/11 of a minute.

 

So the time this happens again in the noon hour is 12:49 and 1/11 of a minute.

 

If we do this for each hour, make sure that we adjust our equations accordingly.

 

I'll say that H is for the hour hand and and M for the minute hand:

 

Now in the 1 o clock hour, for H we'll get 30 + 0.5t because the hour hand will already be at the 1.

 

And for M we'll again get 6t for the first half hour and 360 - 6t for the second half hour.

 

(There might be a way to condense these better, but this is the way I'm thinking to do it for now...)

 

Then to find the time it happens first in the hour 

 

Take M - H = 90

 

And to find the time it happens again in the hour

 

Take 360 - M + H = 90.

 

Do something similar for each hour. 

 

Once you're in the 3 o clock hour, you can adjust how you're doing this again:

 

H = 90 + 0.5t

 

M = 6t 

 

To get the first pass, notice that at t = 0 it's already 90. To get the second pass take M - H = 90 and solve for t.

 

In the 4 o clock hour, you'll have

 

H = 120 + 0.5t

 

M = 6t

 

To get the first pass, take H - M = 90. To get the second pass, take M - H = 90.

 

Let's see if we get what they did in the example they gave in the problem:

 

H - M = 120 - 5.5t = 90

 

Then 5.5t = 30

 

11t = 60

 

t = 5 and 5/11

 

So the time will be 4:05 and 5/11.

 

The second pass is M - H = 90

 

M - H = 5.5t - 120 = 90

 

So 5.5t = 210

 

11t = 420

 

t = 38 and 2/11

 

And the time will be 4:38 and 2/11. 

 

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There might be a smarter way to do this than considering each hour separately, but I feel like this is the clearest way to take care of it at the moment because you can picture what's happening. Maybe someone else will have something quicker...

 

Hope this helps!