Can someone provide a solution for this question ...
Given the hours , minutes and seconds hands calculate the number of right angles the three hands make pairwise with respect to each other during a day... So it asks for the second and hour angle , minute and hour and second and minute
Thanks a lot..
Take two hands: a fast hand that completes $x$ revolutions per day, and a slow hand that completes $y$ revolutions per day. Now rotate the clock backwards, at a rate of $y$ revolutions per day: the slow hand comes to a standstill, and the fast hand slows down to $x-y$ revolutions per day. So the number of times that the hands are at right angles is $2(x-y)$.
The three hands make 2, 24, and 1440 revolutions per day, so the total is: $$2\times(24-2) + 2\times(1440-2) + 2\times(1440-24) = 5752$$