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Let's assume that some triangle is described by its two sides and an angle (which is not between the given two sides however).


Basically for a triangle above only characteristics $c,\ b,\ C$ are known. Geometrically (by using visual imagination) it is clear to me that the resulting triangle is not always unique. Specifically for $C<90 \deg$ it is possible to construct both acute and obtuse triangles. Further complications include that apparently in some cases no triangle can be constructed at all.

What could be an algebraic proof of this non-uniqueness? I guess it would be an unpleasant procedure to try to explicitly derive $A$ value for example but maybe it's easier to show that this characteristic could have multiple solutions?

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It isn't unpleasant at all to determine the values of $A$. What you've expressed in the first paragraph is well known by the congruence theorems. Why do you want to have an algebraic proof? –  Michael Hoppe Oct 16 '13 at 8:49
@MichaelHoppe Now I see that this one is referred to as ASS (angle-side-side) case or donkey theorem. But once again there I can find only neat visual illustrations of the situation in question. I just prefer the algebraic way though. –  Pranasas Oct 16 '13 at 9:06
@Pranasas If I understand correctly, you are looking for a numerical example that has multiple solution. If it is, you are invited to take a look at my answer to problem #485668. –  Mick Dec 19 '13 at 3:48
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1 Answer

up vote 1 down vote accepted

Start with point $A$ and place it in the origin of your coordinate system. Put point $C$ at distance $b$ from $A$. Describe the ray $\vec{CB}$ using $\angle C$. Now you have to intersect a circle of radius $c$ around point $A$ with that ray. Which is a quadratic equation that will in general have two solutions. Algebraically you can see this as two possible values for $a$.

$$ A = \begin{pmatrix}0\\0\end{pmatrix} \\ C = \begin{pmatrix}b\\0\end{pmatrix} \\ \vec{CB} = \left\{\begin{pmatrix}b\\0\end{pmatrix}+ \lambda\begin{pmatrix}-\cos\angle C\\\sin\angle C\end{pmatrix} \;\middle\vert\; \lambda\in\mathbb R\right\} \\ B = \begin{pmatrix}b\\0\end{pmatrix}+ a\begin{pmatrix}-\cos\angle C\\\sin\angle C\end{pmatrix} \\ \lVert A-B\rVert^2 = \lVert B\rVert^2 = (b-a\cos\angle C)^2+(a\sin\angle C)^2 = c^2 \\ a^2-2a\cos\angle C+b^2-c^2 = 0 \\ a_{1,2} = \cos\angle C\pm\sqrt{\cos^2\angle C-b^2+c^2} $$

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I love it! This provided me with a significantly deeper insight than a visual illustration did. –  Pranasas Oct 17 '13 at 11:51
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