Question

Prove that in any triangle ABC,cos A = (b² + c² - a²)/2bc  where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively

Answer 1

Given:

a, b, c are magnitudes of the sides opposite to the vertices A, B, C respectively.

⇒ AB = c, BC = a and CA = b

To Prove:

In triangle ABC,

cos A = (b² + c² - a²)/2bc

Construction: We have constructed a triangle ABC and named the vertices according to the question.

Note the height of the triangle, BD.

If ∠BAD = A

Then, BD = c sin A

Proof:

Here, components of c which are:

c sin A

c cos A

are drawn on the diagram.

Using Pythagoras theorem which says that,

(hypotenuse)² =(perpendicular)² + (base)²

Take ∆BDC, which is a right-angled triangle.

Here,

Hypotenuse = BC

Base = CD

Perpendicular = BD

We get,

(BC)² = (BD)² + (CD)²

⇒ a² = (c sin A)² + (CD)² [∵ from the diagram, BD = c sin A]

⇒ a² = c² sin² A + (b – c cos A)²

[∵ from the diagram, AC = CD + AD

⇒ CD = AC – AD

⇒ CD = b – c cos A]

⇒ a² = c² sin² A + (b² + (-c cos A)² – 2bc cos A) [∵ from algebraic identity, (a – b)² = a² + b² – 2ab]

⇒ a² = c² sin² A + b² + c² cos² A – 2bc cos A

⇒ a² = c² sin² A + c² cos² A + b² – 2bc cos A

⇒ a² = c² (sin² A + cos² A) + b² – 2bc cos A

⇒ a² = c² + b² – 2bc cos A [∵ from trigonometric identity, sin² θ + cos² θ = 1]

⇒ 2bc cos A = c² + b² – a²

⇒ 2bc cos A = b² + c² – a²

Hence, proved.