# How many times are the hands of a clock at $90$ degrees.

How many times are the hands of a clock at right angle in a day?

Initially, I worked this out to be $$2$$ times every hour. The answer came to $$48$$.
However, in the cases of $$3$$ o'clock and $$9$$ o'clock, right angles happen only once.
So the answer came out to be $$44$$.
Is the approach correct?

• glassdoor.com/Interview/… Dec 14, 2013 at 18:08
• In the case of $3$ and $9$ o'clock, its still $2$ times;at $3:00$ sharp and around $3:30$, at $9:00$ sharp and around $9:30$. Dec 14, 2013 at 18:10
• Not quite, @K.Rmth . At $\;3:30\;$ the hours hand already advanced a little towards the $\;4\;$ ...Anyway it is twice, once at $\;3:30\;$ and the other one at some other hour after $\;3:30\;\ldots$ Dec 14, 2013 at 18:43
• Related: Hands of the clock, Revisited. . Oct 27, 2021 at 9:16

Yes, but a more “mathematical” approach might be this: In a 12 hour period, the minute hand makes 12 revolutions while the hour hand makes one. If you switch to a rotating coordinate system in which the hour hand stands still, then the minute hand makes only 11 revolutions, and so it is at right angles with the hour hand 22 times. In a 24 hour day you get 2×22=44.

• Sorry if this should be obvious, but I don't understand how the minute hand makes only 11 rev when the hour hand is still...? Sep 1, 2017 at 13:17
• @Helena That's because we are looking at the minute hand in a coordinate system which itself is rotating, in the same direction as the minute hand, but more slowly. So you have to subtract the one full rotation of the coordinate system from the twelve rotations that the minute hand makes in the usual (non-rotating) coordinate system. Sep 2, 2017 at 11:29

More mathematically, it can be done as :

The minute hand moves 360 degrees in 60 minutes. This means that the angle of the minute hand is given by 6t, where t is number of minutes past midnight.

The hour hand moves 30 degrees in 60 minutes. This means that the angle of the hours hand is given by 0.5t.

The hands start together at midnight. The first time they make a 90 degree angle is when the minute hand has moved 90 degrees further than the hour hand, so this is given by the equation:

6t = 0.5t + 90

5.5t = 90

t = 16 4/11 (16 minutes and 4/11 seconds)

In other words about 16 minutes past midnight.

The next time is when the minutes hand has gained another 180 degrees on the hour hand, and is 90 degrees behind it:

6t = 0.5t + 270

5.5t = 270

t = 49 1/11 (49 minutes and 1/11 seconds)

At about 11 minutes to 1 o'clock.

For every 180 degrees that the minute hand gains on the hour hand there will be one 90 degree angle, so every 49 1/11 - 16 4/11 = 32 8/11 minutes

24 hours is 1440 minutes. 1440/(32 8/11 ) = 44

So every 24 hours there are 44 right angles between minute hand and second hand.

Hope it helps :)

The question reads, "How many times are the hands of a clock at right angle in a day?"

There are three hands on a clock.
The number of occurrences of right angles between the hour and minute hands in a 24-hour period is 44.
The number of occurrences of right angles between the minute and second hands in a 24-hour period is 2832.
The number of occurrences of right angles between the second and hour hands in a 24-hour period is 2876.
Soooo-ooo: 5752.

To the down-voter(s): Is my answer wrong, or do you not like the typography, or what?

• I like this answer and the way you addressed the down voters Aug 28, 2016 at 5:05
• Your answer needs a better explanation. As to why is it 2832 and 2876.
– Alex
Jul 28, 2020 at 10:59

The following C code prints out the times of the day (12hr cycle) that the hands make 90degree angles (to the nearest 3 seconds to avoid the need for floating point). It is indeed 44 times.

The pattern actually repeats itself every 6 hours.

#include <stdio.h>
#include <unistd.h>

void main()
{
int uair=0;
int soicind=0;
int lamhghearr=0;
int uillinn=0;
float ceim=0.0;

for(uair=0; uair<12; uair++){
soicind=0;
for(soicind=0; soicind<60; soicind=soicind+3){
if( ((uillinn >= 897500) & (uillinn <= 902500) ) | ((uillinn >= 2697500) & (uillinn <= 2702500) )){
ceim=(float)(uillinn)/10000;
printf("There are  %6.4f degrees between the hands at %02d:%02d:%02d o' clock\n", ceim, uair, noimead, soicind);
}
lamhghearr=lamhghearr+25;                        /* 0.025 degrees every 3 seconds */
}
}
}
}


This Is situation of 12 hrs. Usually in each hour there are 2 90 deg but these two are exceptions. This repeats again and thus there are 4 exceptions. Since there must be 48 90 deg but excluding 4 exceptions there 48-4 =44 90 deg angles.

• Answer is very exact . And should consider true :) Jun 28, 2017 at 10:27

Correct answer is 22*2. In every hour there are two instances of making 90 degree angle but in case of 3 and 9 there are exceptions. Between 2 and 4 there will be only three right angles because right angle at 2:59 and 3:00 is common and right angle at 8:59 and 9:00 is common. Total will be 22+22 = 44.

Let us think like that: How many times for 12 hours do the two hands meet? 11. You can imagine this: the first time to short hand is between 12 and 1, the second time between 1 and 2, etc. Between two meets they make a right angle exactly twice, this is 11*2 = 22 times right angle per 12h.

Now we have 22*2 = 44 times per 24h. And yes, we can double since at 12pm the two heads are one above the other as in the very start at 12am.

I used Prashant Bhalani's idea from: here

I think the total number of times, clock hands are perpendicular is 22. One of the solution converges to the other, 8:59 ~= 9:00 and 2:59 ~= 3:00, so there are 11 times per 12h.

Gain in angle for 360° = 330° In 24 hours total angle = 24*360° Total angle gain = 24*330° For every 180° relative angle position for the hands of clocks will be straight Hence, (24*330)/180= 44

• This is a pretty old question with a large number of existing answers, several of which give a much more complete explanation than your answer. It might be better to use your time answering unanswered questions. Moreover, I am confused by your answer---how is $360^{\circ} = 330^{\circ}$? May 16, 2018 at 16:37