# The Cosine Law - Proof

Let ABC be a triangle, acute angled in fig. (i), obtuse angled in fig. (ii) and right angled in fig (iii) .

BD is drawn perpendicular to AC from point B in fig(i) and fig(ii).

To prove:

Proof:**1. When A is acute-angled as shown in fig(i),**

CD = AC - AD.......................(i)

In right-angled triangle ABD, by Pythagoras theorem,

BD^{2} = AB^{2} - AD^{2} .................(ii)

Also,

In right-angled triangle BCD, by Pythagoras theorem,

BC^{2} = BD^{2} + CD^{2}

BC^{2} = AB^{2} - AD^{2} + CD^{2} [ **from eqn (ii) **]

BC^{2} = AB^{2} - AD^{2} + (AC - AD)^{2} [ **from eqn (i) **]

BC^{2} = AB^{2} - AD^{2} + AC^{2} + AD^{2} - 2.AC.AD

BC^{2} = AB^{2} + AC^{2} - 2.AC.AD

BC^{2} = AB^{2} + AC^{2} - 2.AB.AC cos A [ from eqn (iii) ]

i.e., **2. **When A is obtuse-angled as shown in fig(ii),

CD = AC + AD..................(i)

In right-angled triangle ABD, by Pythagoras theorem,

BD^{2} = AB^{2} - AD^{2} .................(ii)

Also,

In right-angled triangle BCD, by Pythagoras theorem,

BC^{2} = BD^{2} + CD^{2}

BC^{2} = AB^{2} - AD^{2} + CD^{2} [ from eqn (ii) ]

BC^{2} = AB^{2} - AD^{2} + (AC + AD)^{2} [ from eqn (i) ]

BC^{2} = AB^{2} - AD^{2} + AC^{2} + AD^{2} +2.AC.AD

BC^{2} = AB^{2} + AC^{2} + 2.AC.AD

BC^{2} = AB^{2} + AC^{2} - 2.AB.AC cos A [ from eqn (iii) ]

i.e.,

3. When A is right-angled as shown in fig(iii),

Here, A=90^{0}

Taking 'cos' on both sides,

cos A = cos90^{0} = 0...........(i)

In triangle ABC, by Pythagoras theorem,

BC2 = AB2 + AC2

BC2 = AB2 + AC2 - 2.AB.AC cos A [ **from eqn (i) **]

i.e.,

Hence for all triangles, we have

i.e.,

Hence, proved _{ }

#In a similar manner, the remaining two relations of cos B and cos C can be established