If xcos⁡θ=ycos⁡θ−2π3=zcos⁡θ+2π3 then x+y+z is equal to

If xcosθ=ycosθ2π3=zcosθ+2π3 then x+y+z is equal to

  1. A

    1

  2. B

    0

  3. C

    -1

  4. D

    None of these

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    Solution:

    we have,xcosθ=ycosθ2π3=zcosθ+2π3

    Therefore, each ratio is equal to 

    x+y+zcosθ+cosθ2π3+cosθ+2π3=x+y+zcosθ+2cosθcos2π3=x+y+z0 x+y+z=0

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