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Cos Case Concern (Posted on 2015-12-06) Difficulty: 3 of 5
O is the center of a circle. AB, CD and EF are three parallel chords of this circle having respective lengths 2, 3 and 4.
It is known that ∠AOB = m, ∠COD = n and, ∠EOF = m+n, where m+n < 180o

Determine cos m

No Solution Yet Submitted by K Sengupta    
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Solution Solution | Comment 3 of 4 |

Let the radius of the circle be R.
Applying the law of cosines to triangle AOB
we get 
         cos(m) = 1 - 2^2/(2*R^2)         (1)
So all we need is R^2.
Let G be the point on the smaller arc EF
such that |GF| = |AB| = 2.
   |GF| = 2  ==>  /GOF = /AOB = m
             ==>  /EOG = /EOF - /GOF
             ==>  /EOG = (m+n) - m = n
             ==>  |EG| = 3
Therefore, R is the circumradius of triangle
EFG whose side lengths are 2, 3, and 4.
The circumradius squared of a triangle whose
side lengths are a, b, and c is
                (a*b*c)^2
   R^2 = ------------------------
          16*s*(s-a)*(s-b)*(s-c)
,where s is its semiperimeter.
Plugging in 2, 3, and 4 for a, b, and c we get
   R^2 = 64/15
Plugging this into equation (1) gives
   cos(m) = 17/32 = 0.53125
QED


  Posted by Bractals on 2015-12-06 15:46:04
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