# 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),
In right-angled triangle ABD, by Pythagoras theorem,
BD2 = AB2 - AD2 .................(ii)
Also,

In right-angled triangle BCD, by Pythagoras theorem,
BC2 = BD2 + CD2
BC2 = AB2 - AD2 + CD2                                          [ from eqn (ii) ]
BC2 = AB2 - AD2 + (AC - AD)2                               [ from eqn (i)  ]
BC2 = AB2 + AC2 - 2.AC.AD
BC2 = AB2 + AC2 - 2.AB.AC cos A                       [ from eqn (iii) ]
i.e.,

2. When A is obtuse-angled as shown in fig(ii),
In right-angled triangle ABD, by Pythagoras theorem,
BD2 = AB2 - AD2 .................(ii)
Also,

In right-angled triangle BCD, by Pythagoras theorem,
BC2 = BD2 + CD2 ﻿
BC2 = AB2 - AD2 + CD2                                          [ from eqn (ii) ]
BC2 = AB2 - AD2 + (AC + AD)2                              [ from eqn (i)  ]
BC2 = AB2 + AC2 + 2.AC.AD
BC2 = AB2 + AC2 - 2.AB.AC cos A                       [ from eqn (iii) ]
i.e.,

3. When A is right-angled as shown in fig(iii),
Here, A=900
Taking 'cos' on both sides,
cos A = cos900 = 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