Question

In ΔABC, prove that a² = b² + c² – 2bc cos A.

Answer 1

Let us take the angle A of △ABC in standard position, i.e. A as origin, X-axis along the line AB and the Y-axis perpendicular to the line AB.

In the two figures, ∠A is shown as acute in one and obtuse in the other.

∵ l(AB) = c

∴ B ≡ (c, 0)

Let C ≡ (x, y). Since l(AC) = b, we have

cos  A = `x/b` and sin A = `y/b`

∴ x = b cos A and y = b sin A

∴  C ≡ (b cos A, b sin A)

∴ By the distance formula

a² = BC² = (c – b cos A)² + (0 – b sin A)²

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

= b²(cos²A + sin²A) + c² – 2bc cos A

∴ a² = b² + c² – 2bc cos A.