**In a triangle ABC,**

**a = b cosC + c cosB**

b = c coaA + a cosC

c = a cosB + b cosA

b = c coaA + a cosC

c = a cosB + b cosA

These are known as projection formulea.

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

Labels: Trigonometry