Question

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

Options

• b

• a + b + c

• a

• c

Answer 1

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