If the sides of a triangle $ABC$ whose perimeter is $42$ are in arithmetic progression,its circum-radius is $\frac{65}{8}$ and $B < A < C$,then $\sin A=$

  • A
    $\frac{4}{13}$
  • B
    $\frac{28}{65}$
  • C
    $\frac{56}{65}$
  • D
    $\frac{14}{65}$

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