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 Quadrilateral With A Difference (Posted on 2007-11-08)
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 No Rating

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 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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