In a triangle $ABC$, if $a:b:c = 4:5:6$, then $\frac{1}{4R}[r_1+r_2+r_3] =$

  • A
    $\frac{71}{64}$
  • B
    $\frac{4}{5}$
  • C
    $\frac{81}{84}$
  • D
    $\frac{7}{9}$

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