In a triangle ABC,

a = b cosC + c cosB
b = c coaA + a cosC
c = a cosB + b cosA

These are known as projection formulea.

( To know the the parameters refer here. )



Proof
Let ABC be any of the triangles in the above figures

In fig (i) we have
a= BC = BD+DC.................(a)
But BD/DA= cosB andDC/CA=cosC
==>BD=c cosB
and
DC=b cosC
Putting these values in (a)
a=c cosB+ bcosC

In fig(ii) BC=BD-CD ........................... (b)
Here CD/CA=cos(180-C)=-cosC
==>CD=-bcosC
Also,BD=ccosB
Putting these values in (b)
a= c cosB-(-b cosC)
=c cosB+ b cosC

Other results can also be proved.

Similarly the results can also be proved for Fig(iii) also.


Example:


In any triangle ABC, prove that
(b +c )cos A+(c +a) cos B+ (a + b )cos C = a+ b+ c


Answer


Formula required: Projection Formula


In any triangle ABC, a = b cos C+c cos B

We have L.H.S=(b +c )cos A+(c +a) cos B+ (a + b )cos C

=b cos A+c cos A+c cos B+a cos B+ a cos C+ b cos C

=(b cos C+c cos B) + (a cos C+ c cos A) +(a cos B+b cos A) [ using projection formula]

=a+b+c

=R.H.S