Question

in triangle abc prove that a(bcosC-ccosB)=b^2-c^2

Answer 1

Hello, thanks for asking a question. i will try my best to answer it properly.

law of cosines:

c² = a² + b² - 2ab cos y

converting equation in law of cosines

b^2 = a^2 + c^2 - 2ac cos B
c^2 = a^2 + b^2 - 2ab cos C

Second equation is subtracted from the first one
and the result came

b^2 - c^2 = c^2 - b^2 - 2ac cos B + 2ab cos C

c^2  and - b^2 goes on the other side of the = gives

2b^2 - 2c^2 = -2ac cos B + 2ab cos C

c^2 - b^2 come in brackets and 2 is taken as common gives

 2(b^2 - c^2) = 2a(-c cos B + b cos C)

So b^2 - c^2 = a(b cos C - c cos B)

i hope my answer will help you. i tried my best to solve it in the easiest way. i hope it helped you.

Answer 2

Answer:

b^2 = a^2 + c^2 - 2ac cos B

c^2 = a^2 + b^2 - 2ab cos C

b^2 - c^2 = c^2 - b^2 - 2ac cos B + 2ab cos C

c^2  and - b^2

2b^2 - 2c^2 = -2ac cos B + 2ab cos C

2(b^2 - c^2) = 2a(-c cos B + b cos C)

So b^2 - c^2 = a(b cos C - c cos B)