# Question 5ce04

Jul 22, 2016

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

$\text{44 times}$

#### Explanation:

Take a look at how an analog clock that has ${90}^{\circ}$ 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 ${90}^{\circ}$ 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 ${90}^{\circ}$ 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

$\frac{{360}^{\circ}}{\text{12 intervals" = 30^@"/ interval}}$

An interval here is simply $1$ hour. This implies that the hour hand completes ${30}^{\circ}$ 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 ${360}^{\circ}$ per hour.

This implies that the minute hand completes

${360}^{\circ} - {30}^{\circ} = {330}^{\circ}$

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 ${0}^{\circ}$, i.e. at $12$ o'clock.

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

In other words, the minute hand is "losing" ${30}^{\circ}$, 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 ${90}^{\circ}$ 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 ${90}^{\circ}$ 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 ${330}^{\circ}$ 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.