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If $\Delta ABC$ if $\large\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$ then $C=$

\[\begin {array} {1 1} (1)\;90^{\circ} & \quad (2)\;60^{\circ} \\ (3)\;45^{\circ} & \quad (4)\;30^{\circ} \end {array}\]

1 Answer

(2) $60^{\circ}$
answered Nov 7, 2013 by pady_1
 
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