Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Pythagoras shows us how to find the 3rd side length on a right angled triangle where the two lengths connected by the 90 degrees are known.

Additionally there is a surprisingly short equation that I can't find that gives a third length where any other two lengths are known regardless of the inside angles of the triangles corners. What is this equation?

Is there a similar equation for knowing only the inside angles of two corners and finding the third, if so what is it?

Also are there any other very general, or even more general triangle equations showing relationships between a triangles side lengths and inside angles of the corners given wide ranges of possible just enough information.

share|cite|improve this question
I think we should better call him Pythagoras. And, corners could well be vertices. – user21436 Dec 22 '11 at 16:03
Thanks for the correction on Pythagoras, had I been using the wrong mathematicians name? I'm not that good at names. Also find me somone who when asked what the pointy things between the edges of triangles are replies 'vertices' and I'll make that change. – alan2here Dec 22 '11 at 17:57
up vote 1 down vote accepted

When you know all angles and two of the sides, you can choose between two different rules for finding the third side. Both apply to general triangles:

  • The law of cosines: $a^2+b^2-2ab\cos C=c^2$. This can be thought of as a generalization of Pythagoras' theorem because the cosine term vanishes when $C$ is a right angle.

  • The law of sines: $\frac{\sin C}{c}=\frac{\sin A}{a}$. This can be solved for $c$ to get $c=a\frac{\sin C}{\sin A}$.

If you google for solving triangles you will find numerous walk-throughs of how to find the entire triangle depending on what you already know about it. The first hit for me is here, which looks good.

In general you need three pieces of information, where a "piece" is either the length of a side or an angle. Three pieces also turn out to be sufficient except in the AAA case (you know all angles and no length, so any solution can be scaled by an arbitrary abount and still be a solution) and the SSA case (where you know to sides and one of the angles, but not the angle where the known sides meet; then there can be two different solutions).

share|cite|improve this answer
oh, nice hint! tyvm :¬) I've marked this answer as correct since it was expanded. – alan2here Dec 22 '11 at 16:02

Let $\vec{a}$ and $\vec{b}$ be two vectors corresponding to sides of a triangle. Then the third side is $\vec{c} = \vec{a} - \vec{b}$.

$$ || \vec{c} ||^2 = \langle \vec{c}, \vec{c} \rangle = \langle \vec{a}-\vec{b}, \vec{a}-\vec{b} \rangle = \langle \vec{a}, \vec{a} \rangle + \langle \vec{b}, \vec{b} \rangle -2 \langle \vec{a}, \vec{b} \rangle = | \vec{a} |^2 + |\vec{b} |^2 - 2 | \vec{a} | \cdot | \vec{c} | \cdot \cos \angle(\vec{a}, \vec{b}) $$

share|cite|improve this answer

Given two lengths of a general tringle and the angle betwen them you can calculate the third length by use of the Law of cosines:

$c = \sqrt{a^2 + b^2 - 2ab\cos\gamma},\quad$ for $\, a^2 + b^2 - 2ab\cos\gamma>0$

share|cite|improve this answer
+1 for being the most general and providing a relatively straightforward equation. Does γ represent the angle between the two known lengths? Can this be rearranged in terms of c not squared like this? c=sqrt(a²+b²−2*abcos(γ)) – alan2here Dec 22 '11 at 15:36
Yes, both is correct. – lumbric Dec 22 '11 at 15:42
tyvm. For others reading this, Wolfram style info on the law of cosines – alan2here Dec 22 '11 at 15:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.