triangle solving

Let us consider skew-angled triangles.  If one knows three parts of a triangle, among which at least one side, then the other parts may be calculated by using the law of sines and the law of cosines.  We distinguish four cases:

1. ASA.  Known two angles and one side, e.g. $\alpha$, $\beta$, $c$.  Other parts:

   $$\gamma ={180}^{\circ }-(\alpha +\beta ),a=\frac{c\sin \alpha }{\sin \gamma },b=\frac{c\sin \beta }{\sin \gamma }$$

   

   Figure 1: ASA (angle-side-angle)

2. SSS.  Known all sides $a$, $b$, $c$.  The angles are obtained from

   $$\cos \alpha =\frac{b^2+c^2-a^2}{2bc},\cos \beta =\frac{c^2+a^2-b^2}{2ca},\cos \gamma =\frac{a^2+b^2-c^2}{2ab}.$$

   

   Figure 2: SSS (side-side-side)

3. SAS.  Known two sides and the angle between them, e.g. $b$, $c$, $\alpha$.  Other parts from

   $$a^2=b^2+c^2-2bc\cos \alpha ,\sin \beta =\frac{b\sin \alpha }{a},\sin \gamma =\frac{c\sin \alpha }{a}$$

   

   Figure 3: SAS (side-angle-side)

4. SSA.  Known two sides and the angle of one of them, e.g. $a$, $b$, $\alpha$.  Other parts are gotten from

   $$\sin \beta =\frac{b\sin \alpha }{a},\gamma ={180}^{\circ }-(\alpha +\beta ),c=\frac{a\sin \gamma }{\sin \alpha }.$$

   

   Figure 4: SSA (side-side-angle)

   Since the SSA criterion alone does not prove congruence,it is not surprising that there may not always be a single solutionfor $\beta$ here.In fact, if the first equation gives  $\sin \beta >1$,  then the situation is impossible and the triangle does not exist.  If the equation gives  ,  one gets two distinct values of $\beta$; an acute ${\beta }_1$ and an obtuse  ${\beta }_2={180}^{\circ }-{\beta }_1$.  If in this case  ${\beta }_1>\alpha$,  then there are two different triangles as , but if ${\beta }_1\le \alpha$,  then there is only one triangle.

• http://svn.gold-saucer.org/repos/PlanetMath/TriangleSolving/triangle.mpMetaPost source code for the above diagrams