# Prove that in any triangle ABC, where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively

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.

Then, BD = c sin A

BD = c sin A]

And, 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

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 = 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.

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