Whenever we deal with clock problems, these are some points that come in handy.

1. Every hour is \(\frac{360^o}{12} = 30^o\) in measure
2. Every minute will cause the minute hand to move \(\frac{360^o}{60} = 6^o\). As 12 minutes
pass in every hour, the hour hand will move \(\frac{30^o}{5} = 6^o\) towards the next hour.

Coming to the problem at hand, we have the hour hand standing between 5 and 6.

The hour hand has moved \(\frac{44}{60}*30 = 22^o\) away from 5(which is at \(150^o\)) and is at \(172^o\).
The minute hand is at \(44* 6^o = 264^o\). The angle between the hour and the minute hand
is the difference between the hour and minute hand.

Therefore, the angle between the minute and hour hand is \(264^o - 172^o = 92^o\)(Option D)
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we can use formula
angle = 11/2* min-30*hr
angle = 11/2* 44-30*5 ; 92
IMO D

=>



The minute hand moves \(6°\) every minute and \(<x = 44*6°=264°.\)

The hour hand moves \(0.5°\) every minute and \(<y = 44*0.5°+150 = 172°.\)

Then the angle between the minute hand and the hour hand is \(264°-172°=92°.\)

Therefore, D is the answer.
Answer: D
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