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If Two Sides of a Triangle Are Unequal, Then the Angle Opposite to the Greater Side Is Greater Than Angle Opposite to the Smaller Side

In ABC prov...

Question

In $△ A B C$ prove that
$a^{3} sin (B - C) + b^{3} sin (C - A) + c^{3} sin (A - B) = 0$ .

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Solution

LHS

$= a^{3} sin (B - C)$

$= (2 R sin A)^{3} sin (B - C)$

$= 2 R^{3} \cdot 2 {sin}^{2} A \cdot 2 sin A \cdot sin (B - C)$

$= 2 R^{3} \cdot (1 - cos 2 A) \cdot [cos (A - B + C) - cos (A + B - C)]$

$= 2 R^{3} \cdot (1 - cos 2 A) \cdot [cos (π - 2 B) - cos (π - 2 C)]$

$= 2 R^{3} \cdot (1 - cos 2 A) \cdot (cos 2 C - cos 2 B)$

$= 2 R^{3} \cdot (cos 2 C - cos 2 B - cos 2 A \cdot cos 2 C + cos 2 A \cdot cos 2 B)$

Similarly, the second part

$= 2 R^{3} (cos 2 A - cos 2 C - cos 2 A \cdot cos 2 B + cos 2 B \cdot cos 2 C)$

And the third part

$= 2 R^{3} (cos 2 B - cos 2 A - cos 2 B \cdot cos 2 C + cos 2 A \cdot cos 2 C)$

Adding up all these three parts, we get,

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

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Similar questions

△ A B C

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

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

In any $Δ A B C,$ prove that

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

△ A B C

A, B, C

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

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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