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.


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 $a+b,a-b$: $$a,b = \dfrac{\sqrt{229}+3}{2}, \dfrac{\sqrt{229}-3}{2}.$$

  • $\begingroup$ 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

@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.

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

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