मराठी

In Δ ABC; with usual notations, bBcCB - CbsinB-csinCsin(B - C) = _______. -

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प्रश्न

In Δ ABC; with usual notations, `("b" sin "B" - "c" sin "C")/(sin ("B - C"))` = _______.

पर्याय

  • b

  • a + b + c

  • a

  • c

MCQ
रिकाम्या जागा भरा

उत्तर

In Δ ABC; with usual notations, `("b" sin "B" - "c" sin "C")/(sin ("B - C"))` = a.

Explanation:

We have, `("b" sin "B" - "c" sin "C")/(sin ("B - C"))`

`= ("k" sin "B" sin "B" - "k" sin "C" sin "C")/(sin ("B - C"))`    ...(Using sine rule)

`= ("k"[sin^2 "B" - sin^2 "C"])/(sin ("B - C"))`

`= ("k" sin("B + C") sin ("B - C"))/(sin ("B - C"))`

= k sin (B + C)

= k sin (180° - A)

= k sin A = a

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