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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

591.0. "<><><> 7 o'clock and angle bisection problem <><><>" by 7481::KOSTAS (Wisdom is the child of experience.) Mon Sep 29 1986 11:45

    Hello,
    
       here is the problem:
    
    
          "How long after  7  o'clock will the second hand bisect 
          the angle formed by the other two hands?"
    
    
       Plase note that we are interested in the bisection of the acute
       angle.
    
    
    Enjoy,
    
    Kostas G.
T.RTitleUserPersonal
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591.1Use angles (no trig!)GALLO::EKLUNDDave EklundMon Sep 29 1986 14:0120
    	Let A be the angle (in degrees) through which the second hand
    travels, B the angle for the Minute hand, and C the angle for the
    hour hand.  Then B = 1/60 * A and C = 1/3600 * A.
    
    When bisected by the second hand, the two angles are equal, expressed
    by (note that 7:00 is 210 degrees, 12:00 is 360 degrees):
    
    	A - (210+C) = (360-A) + B
    
    Substituting for B and C from above we get:
    
    	A - (210 + 1/3600 * A) = 360 - A + 1/60 * A
    
    Solving for A we get:
    
    	A=2052000/7139 degrees or 287.+ degrees.
    
    This translates into 7:00:47.8+ for the time.  Incidently, the two
    equal angles are about 77 degrees.