If in a ΔABC, 1a+c+1b+c=3a+b+c then ∠C=

a+b+ca+c+a+b+cb+c=3⇒1+ba+c+1+ab+c=3⇒ab+c+ba+c=1⇒a2+b2−ab=c2⇒∠C=π3

If in a ΔABC, 1a+c+1b+c=3a+b+c then ∠C=

a+b+ca+c+a+b+cb+c=3⇒1+ba+c+1+ab+c=3⇒ab+c+ba+c=1⇒a2+b2−ab=c2⇒∠C=π3

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If in a $Δ A B C, \frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c}$ then $∠ C$ =

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$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{2}$

$\frac{π}{6}$

answer is A.

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$\frac{a + b + c}{a + c} + \frac{a + b + c}{b + c} = 3 \Rightarrow 1 + \frac{b}{a + c} + 1 + \frac{a}{b + c} = 3$

$\Rightarrow \frac{a}{b + c} + \frac{b}{a + c} = 1 \Rightarrow a^{2} + b^{2} - a b = c^{2} \Rightarrow ∠ C = \frac{π}{3}$

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If in a ΔABC, 1a+c+1b+c=3a+b+c then ∠C=

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If in a $Δ A B C, \frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c}$ then $∠ C$ =