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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

Verified by Experts

The correct Answer is:
C
We have, `"ss"(bsinB-c sin C)/(sin(B-C))`
`=(ksinBsinB-ksinCsinC)/(sin(B-C))` (using sine rule)
`=(ksin^(2)B-sin^(2)C)/(sin(B-C))`
`=(ksin(B+C)sin(B-C))/(sin(B-C))`
`=ksin(B+C)=ksin(180^(@)-A)`
`=ksinA=a`
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