For the triangle $A B C$ , prove that
$R r (sin A + sin B + sin C) = Δ$

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Solution

As $\frac{A}{sin A} = \frac{B}{sin B} = \frac{c}{sin C} = 2 R$
$\Rightarrow$ $R r (sin A + sin B + sin C)$
$= R r (\frac{a + b + c}{2 R})$
$= \frac{r}{2} (2 S) = S = Δ$

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A B C

96 s q . u n i t s

R r (sin A + sin B + sin C)

In $△ A B C, R r (sin A + sin B + sin C)$ is equal
to

A B C

R r (sin A + sin B + sin C)

=

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a (\sin B - \sin C) + (\sin C - \sin A) + c (\sin A - \sin B) = 0

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}

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