The area (in square units) of $\triangle ABC$ if $\angle A=75^{\circ}, \angle B=45^{\circ}$ and $a=2(\sqrt{3}+1)$ is

  • A
    $6$
  • B
    $2\sqrt{3}$
  • C
    $6-2\sqrt{3}$
  • D
    $6+2\sqrt{3}$

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