Question

in triangle abc (b-c)sinA+(c-a)sinB(a-b)sinC=0

Answer 1

$\dfrac{a}{\sin\ A}$ = $\dfrac{b}{\sin\ B}$ = $\dfrac{c}{\sin\ C}$ = 0, it is proved.

Step-by-step explanation:

L.H.S. = (b - c)$\sin\ A$ +  (c - a)$\sin\ B$ +  (a - b)$\sin\ C$    ...(1)

We know that,

Sine rule,

$\dfrac{a}{\sin\ A}$ = $\dfrac{b}{\sin\ B}$ = $\dfrac{c}{\sin\ C}$ = k

∴ a = $k\times\ sinA$, b = $k\times\ sinB$ and

c = $k\times\ sinC$

Putting the value of a, b and c in (1), we get

= k{$\sin\ A\sin\ B$ - $\sin\ A\sin\ C$ + $\sin\ B\sin\ C$

- $\sin\ A\sin\ B$ + $\sin\ A\sin\ C$ - $\sin\ A\sin\ C$}

= k × (0)

= 0 = R.H.S.

Hence, it is proved.