In $Δ A B C$ , a $s i n (B - C) + b s i n (C - A) + c s i n (A - B) =$

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

a + b + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a^{2} + b^{2} + c^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 (a^{2} + b^{2} + c^{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0$
In $△ A B C, a = 2 R sin A = 2 R sin (180^{\circ} - (B + C)) = 2 R sin (B + C)$
$a sin (B - C) + b sin (C - A) + c sin (A - B) = \sum a sin (B - C)$
$= 2 R \sum sin (B + C) sin (B - C)$
$= 2 R \sum ({sin}^{2} B - {sin}^{2} C)$ $(∵ sin (A + B) sin (A - B) = {sin}^{2} A - {sin}^{2} B)$
$= 2 R ({sin}^{2} B - {sin}^{2} C + {sin}^{2} C - {sin}^{2} A + {sin}^{2} A - {sin}^{2} B)$
$= 0$

Suggest Corrections

Similar questions

In $△$ ABC, a sin (B - C) + b sin (C - A) + c sin (A-B) =
[ISM Dhanbad 1973]

Δ A B C

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

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

In $△$ ABC, a sin (B - C) + b sin (C - A) + c sin (A-B) =
[ISM Dhanbad 1973]

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

Join BYJU'S Learning Program