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Let $ABC$ be a triangle such that $∠ACB=π/3$ and let a,b,c denote the lengths of the sides opposite to A,B,C respectively. We know that $a-b=3$ and $c=8$. How can I find the other two sides and angles? I tried to solve it with the sine rule and the cosine rule but somehow something is always missing.

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There is actually enough information to solve the problem. By the law of Cosines, $$a^2+b^2-2ab\cos(\pi/3) = c^2 = 64,$$ which is equivalent to $a^2-ab+b^2 = 64$. Since $a-b = 3$, $a^2-2ab+b^2 = 9$. This means $ab = 64-9 = 55$. So $$a^2+2ab+b^2 = (a^2-ab+b^2)+3ab = 64+3\cdot55 = 229,$$ which means $a+b = \sqrt{229}$. We can now find $a,b$ from $a+b,a-b$: $$a,b = \dfrac{\sqrt{229}+3}{2}, \dfrac{\sqrt{229}-3}{2}.$$

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  • $\begingroup$ But the answer in the textbook is a=7, b=4 $\endgroup$ – TanEma May 19 '17 at 18:14
  • $\begingroup$ Oh, okay. Thank you both. $\endgroup$ – TanEma May 19 '17 at 18:17
  • $\begingroup$ if $a = 7, b = 4, c= 8, C$ is very nearly a right angle (it is slightly obtuse), as $4, 7.5, 8.5$ is a right triangle. $\endgroup$ – Doug M May 19 '17 at 18:25
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@TanEma may be I can explain easier. So you just write cosine theorem first, than use data from your task and rewrite i according what you have enter image description here

After you solve equation you will get answer.

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  • $\begingroup$ thank you, i appreciate it $\endgroup$ – TanEma May 19 '17 at 18:43

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