In triangle ABC, prove the following:
$a (\sin B - \sin C) + (\sin C - \sin A) + c (\sin A - \sin B) = 0$

Open in App

Solution

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

Then,

Consider the LHS of the equation $a (\sin B - \sin C) + (\sin C - \sin A) + c (\sin A - \sin B) = 0$ .

$LHS = a (\sin B - \sin C) + b (\sin C - \sin A) + c (s i n A - s i n B) = k \sin A (\sin B - \sin C) + k \sin B (\sin C - \sin A) + k \sin C (s i n A - s i n B) = k \sin A \sin B - k \sin A \sin C + k \sin B \sin C - k \sin B \sin A + k s i n C \sin A - k \sin C \sin B = 0 = RHS Hence proved .$

Suggest Corrections

Similar questions

In any $Δ A B C$ , prove that.

$a (s i n B - s i n C) + b (s i n C - s i n A) + c (s i n A - s i n B) = 0$

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

\frac{a^{2} sin (B - C)}{sin B + sin C} + \frac{b^{2} sin (C - A)}{sin C + sin A} + \frac{c^{2} sin (A - B)}{sin A + sin B} = 0

A, B, C

△ A B C

(s i n A + s i n B) (s i n B + s i n C) (s i n C + s i n A)

Join BYJU'S Learning Program