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What is the number of degrees in the smaller angle formed by the hour and minute hands of a clock at 5:44?

 Sep 1, 2019
 #1
avatar+106535 
+3

The hour hand  will  move at  a rate  of 

 

360°  in 12 hrs

30° in  1 hr

30°/ 60  =  .5° in one minute

 

So  at   5:44.....the hour  hand has  moved   5 (30°) +  44(.5°)  =  [150 + 22]° = 172°   from 12 noon

 

The minute hand  will move  at a rate of 360°/ 60  = 6° per minute

So at 5:44...it will have moved  [44 * 6]°  = 264°  from the top of the hour

 

So...the smaller angle formed by the hands  =  [264 - 172]°  =  92°

 

 

 

cool cool cool

 Sep 1, 2019
 #2
avatar+4330 
+1

You can also use this formula. \(|30h-5.5m|\), where \(h\) denotes the number of hours and \(m\) is the number of minutes. Thus, the answer is \(|150-242|=|-92|=92\) degrees.

 Sep 2, 2019
edited by tertre  Sep 2, 2019
 #3
avatar+23878 
+3

What is the number of degrees in the smaller angle formed by the hour and minute hands of a clock at 5:44?

 

\(\text{Let the angle formed by the minute hand and hour hand in degrees $\mathbf{ \Delta \alpha }$ } \\ \text{Let the time in hours $\mathbf{ t }$ }\)

 

The formula between the two values \(\Delta \alpha\) and \(t\) is:
\(\boxed{~ \Delta \alpha = 330 \cdot t } \)

 

\(\begin{array}{|rcll|} \hline \Delta \alpha &=& 330 \cdot t \quad | \quad t= 5:44 = 5+\dfrac{44}{60}= 5.7\overline{3}\ h \\ &=& 330 \cdot 5.7\overline{3} \\ &=& 1892 \\ &=& 1892 -5\cdot 360 \\ &=& 1892 - 1800 \\ \mathbf{\Delta \alpha} &=& \mathbf{ 92^\circ } \\ \hline \end{array}\)

 

laugh

 Sep 2, 2019
 #4
avatar+106535 
+2

Thanks, tertre and heureka.....those "formulas"  provide a nice shorthand way to solve this type of problem  !!!!

 

 

cool cool cool

CPhill  Sep 2, 2019
 #5
avatar+23878 
+2

Thank you, CPhill !

 

laugh

heureka  Sep 3, 2019

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