In triangle ABC, prove the following:
$a^{2} \sin (B - C) = (b^{2} - c^{2}) \sin A$

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Solution

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

Consider the RHS of the equation $a^{2} \sin (B - C) = (b^{2} - c^{2}) \sin A$ .

$RHS = k^{2} \sin A (\sin^{2} B - \sin^{2} C) = k^{2} \sin A [\sin (B + C) \sin (B - C)] [∵ \sin^{2} B - \sin^{2} C = \sin (B + C) \sin (B - C)] = k^{2} \sin A [\sin (π - A) \sin (B - C)] [∵ A + B + C = π] = k^{2} \sin A [\sin (A) \sin (B - C)] = k^{2} \sin^{2} A \sin (B - C) = a^{2} \sin (B - C) = LHS [∵ a = k \sin A] Hence proved .$

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Similar questions

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

△ A B C

(i) a s i n A - b s i n B = c s i n (A - B)

(i i) \frac{c - b c o s A}{b - c c o s A} = \frac{c o s B}{c o s C}

(i i i) a^{2} s i n (B - C) = (b^{2} - c^{2}) s i n A

(i v) \frac{(a - b)}{(a + b)} = \frac{tan (\frac{A - B}{2})}{t a n (\frac{A + B}{2})}

(v) a c o s B - b c c o s A = (a^{2} - b 2)

$a^{2} s i n (B - C) = (b^{2} - c^{2}) s i n A$

A B C

c^{2} = a^{2} + b^{2} - 2 a b cos C

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