$P r o v e t h a t i n t r i a n g l e A B C : a^{3} sin (B - C) + b^{3} sin (c - a) + c^{3} sin (A - B) = 0$

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Solution

Given the equation

LHS

$a^{3} s i n (B - C) + b^{3} s i n (C - A) + c^{3} s i n (A - B) = 0$

we know that

$\frac{s i n A}{a} = \frac{s i n B}{b} = \frac{s i n C}{c} = K$ (let)

$c o s A = \frac{a^{2} + c^{2} - a^{2}}{2 b c}$

$c o s B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$

$\frac{b^{2} + a^{2} - c^{2}}{2 b a}$

so,

$a^{3} s i n (B - C) = a^{3} [s i n B c o s C - c o s B s i n C]$

= $a^{3} [k b \times \frac{b^{2} + a^{2} - c^{2}}{2 b a} - \frac{a^{2} + c^{2} - b^{2}}{2 a c} \times k c]$

= $k a^{3} \times \frac{b^{2} + a^{2} - c^{2} - a^{2} - c^{2} + b^{2}}{2 a}$

= $k a^{2} (b^{2} - c^{2})$

similarly,

$b^{3} s i n (C - A) = k b^{2} (c^{2} - a^{2})$

$C^{3} s i n (A - B) = k c^{2} (a^{2} - b^{2})$

now, adding all we get,

= $k [a^{2} (b^{2} - c^{2}) + b^{2} (c^{2} - a^{2}) + c^{2} (a^{2} - b^{2})]$

= $k [a^{2} b^{2} - a^{2} c^{2} + b^{2} C^{2} - a^{2} b^{2} + c^{2} a^{2} - c^{2} b^{2}]$
= 0
hence, the RHS is proved.

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Similar questions

a^{3} sin (B - C) + b^{3} sin (C - A) + c^{3} sin (A - B) = 0.

△ A B C

a^{3} sin (B - C) + b^{3} sin (C - A) + c^{3} sin (A - B) = 0

△ A B C

a^{3} sin (B - C) + b^{3} sin (C - A) + c^{3} sin (A - B) = 0

In any $Δ A B C,$ prove that

$a^{3} s i n (B - C) + b^{3} s i n (C - A) + c^{3} s i n (A - B) = 0.$

Δ A B C, a, b, c

A, B, C

a^{3} sin (B - C) + b^{3} sin (C - A) + c^{3} sin (A - B)

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