In a triangle ABC (1) sinA.sinB.sinC = (Δ/(2R^2 )) (2) sinA.sinB.sinC = (r/(2R))(sinA + sinB + sinC) (3) acosA + bcosB + ccosC = ((abc)/(2R^2 )) (4) sinA.sinB.sinC = (R/(2r))(sinA + sinB + sinC)


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Question Number 18501 by Tinkutara last updated on 22/Jul/17

$In a triangle ABC$ $(1) \sin A . \sin B . \sin C = \frac{Δ}{2 R^{2}}$ $(2) \sin A . \sin B . \sin C = \frac{r}{2 R} (\sin A + \sin B + \sin C)$ $(3) a \cos A + b \cos B + c \cos C = \frac{a b c}{2 R^{2}}$ $(4) \sin A . \sin B . \sin C = \frac{R}{2 r} (\sin A + \sin B + \sin C)$
Commented by b.e.h.i.8.3.417@gmail.com last updated on 22/Jul/17

$Δ = \frac{1}{2} a b . s i n C = \frac{1}{2} . 2 R . s i n A . 2 R . s i n B . s i n C =$ $= 2 R^{2} . s i n A . s i n B . s i n C \Rightarrow a n s w e r (1) i s$ $c o r r e c t .$ $Δ = p . r = \frac{a + b + c}{2} . r = R (s i n A + s i n B + s i n C) . r \Rightarrow$ $\Rightarrow 2 R^{2} Π s i n A = R . r Σ s i n A$ $\Rightarrow Π s i n A = \frac{r}{2 R} . (Σ s i n A) \Rightarrow a n s w e r (2) i s$ $c o r r e c t .$ $a n s w e r (3) i s a l s o, c o r r e c t .$ $a n s w e r (4), i s i n c o r r e c t .$
Commented by Tinkutara last updated on 23/Jul/17

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