Question

Question #5ce04

Answer 1

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.

Answer 2

`"44 times"`

Explanation:

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

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.