If $P$ is a point on the altitude $AD$ of the $\triangle ABC$,and $\angle ABP = \frac{2B}{3}$,then $AP$ is equal to

  • A
    $C \sin \frac{B}{3}$
  • B
    $2C \sin \frac{B}{3}$
  • C
    $C \sin \frac{2B}{3}$
  • D
    $2C \sin \frac{2B}{3}$

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