Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
2572
5
avatar+217 

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+130503 
+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+4623 
+1

You can also use this formula. |30h5.5m|, where h denotes the number of hours and m is the number of minutes. Thus, the answer is |150242|=|92|=92 degrees.

 Sep 2, 2019
edited by tertre  Sep 2, 2019
 #3
avatar+26398 
+3

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

 

Let the angle formed by the minute hand and hour hand in degrees Δα Let the time in hours t 

 

The formula between the two values Δα and t is:
 Δα=330t

 

Δα=330t|t=5:44=5+4460=5.7¯3 h=3305.7¯3=1892=18925360=18921800Δα=92

 

laugh

 Sep 2, 2019
 #4
avatar+130503 
+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+26398 
+2

Thank you, CPhill !

 

laugh

heureka  Sep 3, 2019

3 Online Users

avatar