Answer :

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,

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

[∵ in ∆BAD

⇒ BD = c sin A]

And, AD = c cos A

[∵ in ∆BAD

⇒ AD = c cos 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}

⇒ a^{2} = (c sin A)^{2} + (CD)^{2} [∵ from the diagram, BD = c sin A]

⇒ a^{2} = c^{2} sin^{2} A + (b – c cos A)^{2}

[∵ from the diagram, AC = CD + AD

⇒ CD = AC – AD

⇒ CD = b – c cos A]

⇒ a^{2} = c^{2} sin^{2} A + (b^{2} + (-c cos A)^{2} – 2bc cos A) [∵ from algebraic identity, (a – b)^{2} = a^{2} + b^{2} – 2ab]

⇒ a^{2} = c^{2} sin^{2} A + b^{2} + c^{2} cos^{2} A – 2bc cos A

⇒ a^{2} = c^{2} sin^{2} A + c^{2} cos^{2} A + b^{2} – 2bc cos A

⇒ a^{2} = c^{2} (sin^{2} A + cos^{2} A) + b^{2} – 2bc cos A

⇒ a^{2} = c^{2} + b^{2} – 2bc cos A [∵ from trigonometric identity, sin^{2} θ + cos^{2} θ = 1]

⇒ 2bc cos A = c^{2} + b^{2} – a^{2}

⇒ 2bc cos A = b^{2} + c^{2} – a^{2}

Hence, proved.

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