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Quadrilateral With A Difference (Posted on 2007-11-08) Difficulty: 3 of 5
EFGH is a convex quadrilateral having diagonals EG and FH. It is known that EF = GH, while EH/FG = √3 + 1 and Angle GHE - Angle HEF = 30o.

Determine the value of (Angle EFG - Angle FGH).

See The Solution Submitted by K Sengupta    
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Solution Solution | Comment 2 of 3 |

Using the law of cosines,
  |EG|^2 = |EH|^2 + |HG|^2 - 2|EH||HG|cos(EHG)
         = |EH|^2 + |HG|^2 - 2|EH||HG|cos(t+15)
         = |EF|^2 + |FG|^2 - 2|EF||FG|cos(EFG)
         = |EF|^2 + |FG|^2 - 2|EF||FG|cos(180-[t-x])
  |FH|^2 = |FE|^2 + |EH|^2 - 2|FE||EH|cos(FEH)
         = |FE|^2 + |EH|^2 - 2|FE||EH|cos(t-15)
         = |HG|^2 + (sqrt(3)+1)^2*|EH|^2
           - 2(sqrt(3)+1)*|HG||EH|cos(t-15)
         = |FG|^2 + |GH|^2 - 2|FG||GH|cos(FGH)
         = |FG|^2 + |GH|^2 - 2|FG||GH|cos(180-[t+x])
  ,where 2x = Angle EFG - Angle FGH.
Combining the above two equations we get
  cos(t-x) - cos(t+x) = 
     (sqrt(3)+1)[cos(t-15) - cos(t+15)]
          or
  sin(x) = (sqrt(3)+1)sin(15)
          or
  sin(x)^2 = (sqrt(3)+1)^2*sin(15)^2
           = 2(sqrt(3)+2)*[1-cos(30)]/2
           = 1/2
Therefore, Angle EFG - Angle FGH = 90 degrees.
 

  Posted by Bractals on 2007-11-08 14:20:06
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