Q.

If  xcosθ=ycos[θ-(2π3)]=zcos[θ+(2π3)],  then x+y+z is equal to


Ans.

(0)

We have  xcosθ=ycos[θ-(2π3)]=zcos[θ+(2π3)]

Therefore, each ratio is equal to

x+y+zcosθ+cos[θ-(2π3)]+cos[θ+(2π3)]

         =x+y+zcosθ+2cosθcos(2π3)

          =x+y+z0

⇒x+y+z=0