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In a triangle \(A B C, \frac{a \cos A+b \cos B+c \cos C}{a+b+c}\) is equal to
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Verified Answer
The correct answer is:
\(\frac{\mathrm{r}}{\mathrm{R}}\)
\(\begin{aligned}
& \text {We have, } \frac{a \cos A+b \cos B+c \cos C}{a+b+c} \\
& =\frac{R(\sin 2 A+\sin 2 B+\sin 2 C)}{2 R(\sin A+\sin B+\sin C)} \\
& =\frac{4 \sin A \sin B \sin C}{2\left(4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\right)}: \\
& =4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{r}{R}
\end{aligned}\)
& \text {We have, } \frac{a \cos A+b \cos B+c \cos C}{a+b+c} \\
& =\frac{R(\sin 2 A+\sin 2 B+\sin 2 C)}{2 R(\sin A+\sin B+\sin C)} \\
& =\frac{4 \sin A \sin B \sin C}{2\left(4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\right)}: \\
& =4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{r}{R}
\end{aligned}\)
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