In $\triangle ABC$,if $a=2(\sqrt{3}+1)$,$B=45^{\circ}$ and $C=60^{\circ}$,then the area (in sq.units) of that triangle is

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

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