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If, in $\Delta\; ABC,\large\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$ then the angle $C=$

\[(a)\;30^{\circ} \quad (b)\;45^{\circ} \quad (c)\;60^{\circ} \quad (d)\;90^{\circ} \]

1 Answer

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