In ABC- a sin B - C - b sin C - A - c sin A - B -ISM Dhanbad 1973-A- a2-b2-c2B- 2a2-b2-c2C- 0D- a - b - c

The correct option is C 0 a sin (B C) + b sin(C A)+c sin(A B) = k(∑ A sin(B C) = k∑ sin(B + C) sin(B C) =k = {∑1/2 (cos 2C cos 2B) } = 0. Note: Studen ...

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Standard XII

Mathematics

Sine Rule

In ABC, a sin...

Question

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

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a+b+c

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

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

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Solution

The correct option is A
0

a sin (B-C) + b sin(C-A)+c sin(A-B)
= k( $\sum$ A sin(B - C) = k{ $\sum$ sin(B + C) sin(B - C)}
$= k = {\sum \frac{1}{2} (c o s 2 C - c o s 2 B)} = 0.$

Note: Students should note here that most of the expressions containing the cyclic factor associating with ‘–’ reduces to 0.

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Similar questions

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

Q. Prove that

\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}}

Find:
$(a + b + c) \times (a + b - c) =$

Q. If

a, b

and

c

are in continued proportion, then

\frac{a}{c} =

_____.

Q. In

△ A B C, \frac{s i n (A - B)}{s i n (A + B)} =

[MP PET 1986]

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

The correct option is A 0 a sin (B-C) + b sin(C-A)+c sin(A-B) = k(∑ A sin(B - C) = k{∑ sin(B + C) sin(B - C)} =k={∑12(cos 2C−cos 2B)}=0. Note: Students should note here that most of the expressions containing the cyclic factor associating with ‘–’ reduces to 0.

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