Question

Prove that in any triangle ABC, cos A = `("b"^2 + "c"^2 - "a"^2)/(2"bc")`, where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.

Answer 1

Here, in the given figure, the components of c are c cos A and c sin A.

∴ `vec"CD"` = b – c cos A

In ΔBDC,

a² = CD² + BD²

⇒ a² = (b – c cos A)² + (c sin A)²

⇒ a² = b² + c² cos²A – 2bc cos A + c² sin²A

⇒ a² = b² + c² (cos²A + sin²A) – 2bc cos A

⇒ a² = b² + c² – 2bc cos A

⇒ 2bc cos A = b² + c² – a²

∴ cos A = `("b"^2 + "c"^2 - "a"^2)/(2"bc")`

Hence Proved.