# general triangle angles and lengths

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.

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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

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).

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oh, nice hint! mathsisfun.com/algebra/trig-solving-triangles.html 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 $\vec{c} = \vec{a} - \vec{b}$.

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

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