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In `DeltaABC`, with usual notations, `(bsinB-c sin C)/(sin(B-C))=`. . . .A. bB. `a+b+c`C. aD. c |
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Answer» Correct Answer - 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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