 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Given Ratio, Find Sum (Posted on 2008-01-21) For a triangle ABC, Area/R=4. Find acosA+bcosB+ccosC.
Note:R is circumradius of triangle ABC.

 See The Solution Submitted by Praneeth Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 2 of 5 | `Let O be the circumcenter of triangle ABC.`
`       Area(ABC)     Area(AOB) + Area(BOC) + Area(COA)  4 = ----------- = -----------------------------------           R                         R`
`       R^2*sin(AOB)/2 + R^2*sin(BOC)/2 + R^2*sin(COA)/2    = --------------------------------------------------                               R`
`    = R*(sin(A)*cos(A) + sin(B)*cos(B) + sin(C)*cos(C))       = R*([a/2R]*cos(A) + [b/2R]*cos(B) + [c/2R]*cos(C))`
`                   or`
`    a*cos(A) + b*cos(B) + c*cos(C) = 8`
` `

 Posted by Bractals on 2008-01-21 11:57:33 Please log in:
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