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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

479.0. "What is the radius of the circle?" by KEEPER::KOSTAS (Kostas G. Gavrielidis <o.o> ) Wed Apr 30 1986 22:55

    Hello,
    
         Here is the problem:
    
         "From a point 4 inches outside a circle, a tnagent  12  inches
         long touches the circle, and subtends an angle of  30 degrees
         at the center of the circle.
    
         What is the radius of the circle?"
    
    Enjoy,
    
    Kostas G.
T.RTitleUserPersonal
Name		DateLines
479.1c'mon, these are too easy.METOO::YARBROUGHThu May 01 1986 10:043
    In a 30 degree right triangle the opposite side is half the hypotenuse
    so the hypotenuse is 24. 4 inches is outside the circle so the radius
    is 20. (Yawn.)
479.2Does a picture help?KEEPER::KOSTASKostas G. Gavrielidis <o.o> Thu May 01 1986 23:4731
    re. .1
    
          well no,
    
          by your results  12^2 + 20^2 should equal 24^2.
    
          But it does not. 144 + 400 <> 556.
    
    Maybe if I gave a picture ....
         
                                   |\ <----  tangent touches the circle
                                  /|*\       the angle with R should
                                 / | *\          be 90 degrees.
                                /  | * \
                           R   /   |  * \  12
                              /    |  *  \
                             /   x |   *  \  <----- a tangent 12 inches
                            /      |   *   \
                           /)30    |    *   \
                          /__)_____|____*____\ 
                                  R       4  
                         ^                    ^
                         |                    |
                     center of             a point outside the circle
                     the circle            4 inches a way  
                               
    
    I hope this helps.
    
    Kostas G.
479.3Maybe I _was_ awake then!NMGV01::ASKSIMONDon&#039;t upset the &#039;Deus ex Machina&#039;Fri May 02 1986 06:3226
  I sympathise with .1.  As I see it you have a conflict of
  information.  I did not reply at first because I thought I
  must have misunderstood,  until I saw the diagram,  which leads
  to the impossible triangle I had first envisaged which your
  diagram confirms.

  Because the 12 inch line is a tangent you have 90 degrees
  between it and R.  This creates a right angled triangle between
  the three points (i.e. not including where the R+4 line
  meets the edge of the circle).  You have a hypotenuse of R+4 and an
  angle at the centre of 30 degrees.  This provides two possible
  solutions but with a different answer.  Firstly,  you have
  
  	tan(30) = 12/R	(perpendicular over base)
  
  and secondly
  
  	sin(30) = 12/(R+4)	(perpendicular over hypotenuse)
  
  The first equation gives R=20.78 and the second R=20.  If you want
  the other answer,  i.e. approx. 20.78,  then you should not have
  said "4 inches from" but "somewhere outside the circle"  the
  remaining information would then have been sufficient to arrive at
  the answer by using the above tangent equation.  I must still
  conditionally agree with .1,  though.  I.e. that it would then
  indeed be too easy.
479.4Some history on this ...KEEPER::KOSTASKostas G. Gavrielidis &lt;o.o&gt; Mon May 05 1986 10:1316
    re. .1
    
          The interest on this problem arises from the fact that it
          once (1951) appeared on a paper set for a competition 
          participated in by a large number of schools near Boston.
    
    
    The trouble with it is that it is over-specified.
    
    The three numbers:  4,  12,  and 30,  form an impossible combination
    of data.
    
    If one ignores one of these three numbers, there are three different
    solutions that can be obtained.