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In DeltaABC, with usual notations, (bsin...

In `DeltaABC`, with usual notations, `(bsinB-c sin C)/(sin(B-C))=`. . . .

A

b

B

`a+b+c`

C

a

D

c

Text Solution

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The correct Answer is:
C
We have, `s s (b "sin" B - c "sin" C)/("sin" (B - C))`
`(k "sin" B "sin" B - k "sin" C "sin" C)/("sin"(B - C))`
`(k"sin"^(2) B - "sin"^(2)C)/("sin" (B - C))`
`= k sin (B + C)`
`= k "sin" (180^(@) - A)`
`= K "sin" A = a`
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