Prove that $(b + c) cos A + (c + a) cos B + (a + b) cos C = 2 s .$

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Solution

L.H.S. $= (b + c) cos A + (c + a) cos B + (a + b) cos C$
Using projection rule $c = a cos B + b cos A$
$= (b cos A + a cos B) + (c cos A + a cos C) + (b cos C + c cos B)$
$= c + b + a = 2 s$

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Prove that (b+c)cosA+(c+a)cosB+(a+b)cosC=2s.

L.H.S.=(b+c)cosA+(c+a)cosB+(a+b)cosC Using projection rule c=acosB+bcosA=(bcosA+acosB)+(ccosA+acosC)+(bcosC+ccosB)=c+b+a=2s

Prove that $(b + c) cos A + (c + a) cos B + (a + b) cos C = 2 s .$

L.H.S. $= (b + c) cos A + (c + a) cos B + (a + b) cos C$
Using projection rule $c = a cos B + b cos A$
$= (b cos A + a cos B) + (c cos A + a cos C) + (b cos C + c cos B)$
$= c + b + a = 2 s$