How to solve this $\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$. Given that $\angle C=60°$. Please answer as soon as possible. Tomorrow is my test
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2Traditional Comment : What have you tried? We don't generally answer those questions which don't show any attempt from asker's side. – Jaideep Khare Jun 14 '17 at 13:37
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I took LCM of both and then added it. Then I used cosine rule but nothing worked out. – Muhammad Ali Zafar Jun 14 '17 at 13:39
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Please use MathJax for writing math on this site, or at the very least parentheses. Right now I can't tell whether you mean $\frac 3a + b+c$ or $\frac3{a+b+c}$. – Arthur Jun 14 '17 at 13:40
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use that $$\sin(\alpha)=\frac{a}{2R}$$ etc – Dr. Sonnhard Graubner Jun 14 '17 at 13:40
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Authur i mean the second one. That is what you have written after or. – Muhammad Ali Zafar Jun 14 '17 at 13:43
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@MuhammadAliZafar Can you specify using brackets and edit accordingly? – Jaideep Khare Jun 14 '17 at 13:49
2 Answers
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$$\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\iff$$ $$(a+b+c)(a+b+2c)=3(a+c)(b+c)\iff$$ $$c^2=a^2+b^2-ab=a^2+b^2-2ab\cos(60^o)$$
farruhota
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the left side of your equation is given by $$\left( \sin \left( \alpha \right) +1/2\,\sqrt {3} \right) ^{-1}+ \left( \sin \left( \pi/3+\alpha \right) +1/2\,\sqrt {3} \right) ^{-1} $$ and the right side $$3\, \left( \sin \left( \alpha \right) +\sin \left( \pi/3+\alpha \right) +1/2\,\sqrt {3} \right) ^{-1} $$ and the Difference is $$4\, \left( \sin \left( \alpha \right) \right) ^{2}-4\,\sin \left( \pi /3+\alpha \right) \sin \left( \alpha \right) -3+4\, \left( \sin \left( \pi/3+\alpha \right) \right) ^{2} $$ simplifying this we obtain $$4\sin(\alpha)^2-4\sin((1/3)\pi+\alpha)\sin(\alpha)-3+4\sin((1/3)\pi+\alpha)^2$$ simplifying this we get $$3\, \left( \sin \left( \alpha \right) \right) ^{2}-3+3\, \left( \cos \left( \alpha \right) \right) ^{2} $$ and this is Zero.
Dr. Sonnhard Graubner
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