frac{acos A + bcos B + ccos C}{a + b + c} = | vedclass

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(B) We know that $a = 2R\sin A$,$b = 2R\sin B$,and $c = 2R\sin C$.Substituting these in the expression:$\frac{a\cos A + b\cos B + c\cos C}{a + b + c}

Vedclass · Accepted Answer

$r/R$

Vedclass · Answer

(B) We know that $a = 2R\sin A$,$b = 2R\sin B$,and $c = 2R\sin C$.Substituting these in the expression:$\frac{a\cos A + b\cos B + c\cos C}{a + b + c} = \frac{2R\sin A \cos A + 2R\sin B \cos B + 2R\sin C \cos C}{2R\sin A + 2R\sin B + 2R\sin C}$$= \frac{\sin 2A + \sin 2B + \sin 2C}{2(\sin A + \sin B + \sin C)}$Using the identity $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$ and $\sin A + \sin B + \sin C = 4\cos(A/2)\cos(B/2)\cos(C/2)$:$= \frac{4\sin A \sin B \sin C}{2(4\cos(A/2)\cos(B/2)\cos(C/2))}$$= \frac{4(2\sin(A/2)\cos(A/2))(2\sin(B/2)\cos(B/2))(2\sin(C/2)\cos(C/2))}{8\cos(A/2)\cos(B/2)\cos(C/2)}$$= 8\sin(A/2)\sin(B/2)\sin(C/2)$Since $r = 4R\sin(A/2)\sin(B/2)\sin(C/2)$,we have $4\sin(A/2)\sin(B/2)\sin(C/2) = r/R$.Thus,the expression equals $r/R$.

$\frac{a\cos A + b\cos B + c\cos C}{a + b + c} = $

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