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# Help with Geometry Questions!

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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
+103148
+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°

Sep 1, 2019
#2
+4322
+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
+23086
+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:
$$\boxed{~ \Delta \alpha = 330 \cdot t }$$

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

Sep 2, 2019
#4
+103148
+2

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

CPhill  Sep 2, 2019
#5
+23086
+2

Thank you, CPhill !

heureka  Sep 3, 2019