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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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In $\Delta ABC$ if $\large\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$ then $c$ is equal to

$(a)\;90^{\large\circ}\qquad(b)\;60^{\large\circ}\qquad(c)\;45^{\large\circ}\qquad(d)\;30^{\large\circ}$

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1 Answer

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$\large\frac{a}{b+c}+\frac{b}{c+a}$$=1$
$a(c+a)+b(b+c)=(b+c)(c+a)$
$a^2+ac+b^2+bc=bc+ab+c^2+ac$
$a^2+b^2-c^2=ab$
$\cos C=\large\frac{a^2+b^2-c^2}{2ab}$
$\qquad\;=\large\frac{ab}{2ab}$
$\qquad\;=\large\frac{1}{2}$
$C=60^{\large\circ}$
Hence (b) is the correct answer.
answered Oct 10, 2013 by sreemathi.v
 

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