For a triangle ABC, Area/R=4. Find
acosA+bcosB+ccosC.
Note:R is circumradius of triangle ABC.
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
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Posted by Bractals
on 2008-01-21 11:57:33 |
Solution
