in what duration dies the hour hand and minute hand coincide
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So, time = angular distance/angular relative speed = 360/ (11/2) = 720/11 minutes. In every 720/11 minute, the two hands overlap. So, the number of times the two hands overlap will be equal to number of such minutes in 24 hour duration (24*60 minutes).
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hey mate ur ans
Take two variables for hour “h1 and h2” and then find the angle “theta” [theta = (30 * h1)] and then divide it by “11” to find the time in minute(m). We are dividing with 11, because the hands of a clock coincide 11s time in 12 hours and 22 times in 24 hours :
Formula : This can be calculated using the formula for time h1 to h2 means [11m / 2 – 30 (h1)]
this implies [ m = ((30 * h1) * 2) / 11 ] ]
[ m = (theta * 2) / 11 ]
where [theta = (30 * h1) ]
where A and B are hours i.e if given hour is (2, 3) then A = 2 and B = 3 .
Take two variables for hour “h1 and h2” and then find the angle “theta” [theta = (30 * h1)] and then divide it by “11” to find the time in minute(m). We are dividing with 11, because the hands of a clock coincide 11s time in 12 hours and 22 times in 24 hours :
Formula : This can be calculated using the formula for time h1 to h2 means [11m / 2 – 30 (h1)]
this implies [ m = ((30 * h1) * 2) / 11 ] ]
[ m = (theta * 2) / 11 ]
where [theta = (30 * h1) ]
where A and B are hours i.e if given hour is (2, 3) then A = 2 and B = 3 .
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