Question #5ce04

2 Answers
Jul 22, 2016

Answer:

48 times

Explanation:

In one hour, the clock's short and long hands form a 90° angle 2 times.There are 24 hours in a day. Hence, we have 2x24=48.
Therefore, the clock's short and long hand form 48 90° angles in one day.

Jul 23, 2016

Answer:

#"44 times"#

Explanation:

Take a look at how an analog clock that has #90^@# between the hour hand and the minute hand looks like

https://play.google.com/store/apps/details?id=org.artsplanet.android.simpleanalogclock

Now, the common approach here is to say that in #1# hour, the hour hand and the minute hand are perpendicular to each other #2# times.

This would lead to the conclusion that in #12# hours, the two hands are at #90^@# a total of

#12 color(red)(cancel(color(black)("hours"))) * "2 times"/(1color(red)(cancel(color(black)("hour")))) = "24 times"#

Consequently, you would say that in #24# hours, the two hands are at a #90^@# angle

#24 color(red)(cancel(color(black)("hours"))) * "24 times"/(12color(red)(cancel(color(black)("hours")))) = "48 times"#

This is not the correct answer. Here's why.

As you can see, the circle is divided into #12# equal intervals, each measuring

#(360^@)/"12 intervals" = 30^@"/ interval"#

An interval here is simply #1# hour. This implies that the hour hand completes #30^@# per hour. Now, you know that the minute hand completes a full revolution every hour.

This is equivalent to saying that the minute hand completes #360^@# per hour.

This implies that the minute hand completes

#360^@ - 30^@ = 330^@#

in one hour relative to the hour hand. Here's where the tricky part comes in. Let's assume that you're going to keep the hour hand locked in place at #0^@#, i.e. at #12# o'clock.

In this case, the minute hand will only complete #330^@# in one hour because it's moving #330^@# relative to a hand that is stuck in place.

In other words, the minute hand is "losing" #30^@#, the equivalent of #5# minutes, for every passing hour, i.e. it doesn't complete a full revolution in #1# hour when the hour hand is kept in place.

In #12# hours, this will amount to

#12 color(red)(cancel(color(black)("hours"))) * (30^@"lost")/(1color(red)(cancel(color(black)("hour")))) = 360^@"lost"#

This means that in #12# hours, the minute hand only completes #11# full revolutions. Since the hands are at a #90^@# angle twice per hour, you will have

#overbrace(11color(red)(cancel(color(black)("hours"))))^(color(blue)("the equivalent of 12 hours when both hands are moving")) * "2 times"/(1color(red)(cancel(color(black)("hour")))) = "22 times"#

This implies that in #24# hours, the two hands are at a #90^@# angle a total of

#24 color(red)(cancel(color(black)("hours"))) * "22 times"/(12color(red)(cancel(color(black)("hours")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("44 times")color(white)(a/a)|)))#

So, to sum this up, when both hands are moving, the minute hand completes #12# full revolutions in #12# hours.

When the hour hand is kept in place, the minute hand only completes #11# full revolutions in #12# hours. That is the case because it's moving at #330^@# relative to the hour hand, which is being kept in place.

As a result, two perpendicular positions will be "lost" every #12# hours, which is equivalent to saying that #4# perpendicular positions are "lost" every #24# hours.

This is why the two hands are not perpendicular to each other #48# times in #24# hours, but only #44# times in #24# hours.