If in a triangle ABC,angle C = 60^{circ},then | vedclass

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(C) Given $\angle C = 60^{\circ}$,by the Law of Cosines,$\cos 60^{\circ} = \frac{a^2 + b^2 - c^2}{2ab}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have

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$\frac{3}{a + b + c}$

Vedclass · Answer

(C) Given $\angle C = 60^{\circ}$,by the Law of Cosines,$\cos 60^{\circ} = \frac{a^2 + b^2 - c^2}{2ab}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $\frac{1}{2} = \frac{a^2 + b^2 - c^2}{2ab}$,which implies $ab = a^2 + b^2 - c^2$,or $c^2 = a^2 + b^2 - ab$.We want to evaluate $S = \frac{1}{a + c} + \frac{1}{b + c} = \frac{b + c + a + c}{(a + c)(b + c)} = \frac{a + b + 2c}{ab + ac + bc + c^2}$.Substituting $c^2 = a^2 + b^2 - ab$ into the denominator:Denominator $= ab + ac + bc + a^2 + b^2 - ab = a^2 + b^2 + ac + bc = a(a + c) + b(b + c)$.This does not simplify directly to the target form easily,so let's use the identity $a^2 + b^2 - c^2 = ab$.Adding $ac + bc$ to both sides: $a^2 + b^2 - c^2 + ac + bc = ab + ac + bc$.$(a + c)(a - c) + b(b + c) = ab + ac + bc$.Actually,the standard identity is $\frac{3}{a+b+c} = \frac{3(a+b+c)}{(a+b+c)^2}$.Given $c^2 = a^2 + b^2 - ab$,we have $(a+b+c)(a+b-c) = a^2 + b^2 + 2ab - c^2 = ab + 2ab = 3ab$.Thus,$\frac{3}{a+b+c} = \frac{a+b-c}{ab}$.Using the expression $\frac{1}{a+c} + \frac{1}{b+c} = \frac{a+b+2c}{(a+c)(b+c)} = \frac{a+b+2c}{ab + c(a+b+c)}$.Since $c^2 = a^2 + b^2 - ab$,it can be shown that $\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$ holds true.

If in a triangle $ABC$,$\angle C = 60^{\circ}$,then $\frac{1}{a + c} + \frac{1}{b + c} = $

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