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 Acute triangle, Trigonometric function! (Posted on 2007-10-13)
Prove that in an acute triangle sin(A) + sin(B) + sin(C) > 2

 See The Solution Submitted by Chesca Ciprian Rating: 4.0000 (1 votes)

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 Solution Comment 9 of 9 |
`Let a, b, and c be the lengths of sides BC,CA, and AB respectively. Let s be thesemiperimeter (a+b+c)/2. Let O be thecircumcircle of triangle ABC and R thecircumradius.`
`Let point D be a point on side BC such that|AC|+|CD| = |AB|+|BD| = s. Let M be themidpoint of line segment AD. Let N be thecircle with center M and radius s/2.`
`Triangle ABC lies inside circle N since`
`  |AM| = |DM| = |AD|/2 < (|AB|+|BD|)/2 = s/2`
`and points B and C lie on an elipse withcenter M and major axis s/2.`
`If more than 180 degrees of arc of circle Olies outside of circle N, then angle A wouldbe greater than 90 degrees. Therefore, R < s/2.`
`From the rule of sines,`
`      a          b          c    -------- = -------- = -------- = 2R   sin(A)     sin(B)     sin(C)`
`Therefore,`
`                              a+b+c     s  sin(A) + sin(B) + sin(C) = ------- = --- > 2                               2R       R`
` `

 Posted by Bractals on 2007-10-15 05:41:03

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