Question

In any triangle abc prove that a(b cos C - c cos B) = b^2 - c^2.

Answer 1

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)

Answer 2

Let the angles of the triangle be A , B and C and let the sides of the triangle be a , b and c .

By using the Cosines Law we get :

a² + c² - 2 ac cos B = b² --------(1)

Again by using the Cosines Law we get :

a² + b² - 2 ab cos C = c² ---------(2)

Subtract equation (2) from (1) :

⇒ c² - 2 ac cos B - b² + 2 ab cos C = b² - c²

Transpose c² and -b² to the other side :

⇒ 2 ab cos C - 2 ac cos B = b² - c² + b² - c²

⇒ 2 ab cos C - 2 ab cos B = 2 b² - 2 c²

Take 2 as common on both sides :

⇒ 2 ( ab cos C - ac cos B ) = 2 ( b² - c² )

⇒ ab cos C - ac cos B = b² - c²

Take a as common in the left side :

⇒ a ( b cos C - c cos B ) = b² - c²

Hence it is proved .