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Given Ratio, Find Sum (Posted on 2008-01-21) Difficulty: 2 of 5
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)

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Solution 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
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