# Find the sides of a triangle given one side, the difference of two and one angle

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.

## 2 Answers

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

• But the answer in the textbook is a=7, b=4 – TanEma May 19 '17 at 18:14
• Oh, okay. Thank you both. – TanEma May 19 '17 at 18:17
• 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. – 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 After you solve equation you will get answer.

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