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Is it possible to find the angles of a triangle if I only have its sides? If so, how can I achieve this?

Regarding my knowledge of triangles: Whilst I was taught trigonometry a few years ago, I cannot for the life of me remember how to do things like use SOHCAHTOA to figure out the length of a side given an angle and a side. I know it's possible and if that were my problem I would continue searching the internet for a solution, but I gather finding an angle without knowing any of the angles is more difficult.

Any input is appreciated, and I thank you for your time in advance.

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If you want to learn it then you need to study the Law of Cosines. If you need a formula then it is $\cos A={{b^2+c^2-a^2}\over {2ab}}$ –  Maesumi Feb 2 '13 at 17:56
is that (2*a)*b or 2*(b*a)? –  Pharap Feb 2 '13 at 18:35
Sorry to correct the formula I wrote it is $\cos(A)={{b^2+c^2-a^2}\over {2*b*c}}$. In the denominator either $(2*b)*c$ or $2*(b*c)$ will be same. –  Maesumi Feb 2 '13 at 19:36

1 Answer 1

Use the Cosine Law.

Let $\triangle ABC$ have sides $a$, $b$, and $c$. We are using the usual convention that the length of the side opposite vertex $A$ is called $a$, and so on.

Let $\theta=\angle C$. Then the Cosine Law says that $$c^2=a^2+b^2-2ab\cos \theta.$$ Since we know $a$, $b$, and $c$, we can use the above formula to calculate $\cos\theta$. Then we can use the $\cos^{-1}$ button on the calculator to find $\theta$ to excellent accuracy.

We can use the Cosine Law three times to get the three angles. But we only need to do the calculation for two of the angles: If we have them, the third can be easily found.

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Once you have one angle, it’s a bit quicker to use the Law of Sines for the other angles. By the way, I recommend to high-school students to find the largest angle first, for then the ambiguity in using L of S for angles does not occur. –  Lubin May 1 '14 at 17:20

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