# 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. 1.

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

 $\gamma=180^{\circ}\!-\!(\alpha\!+\!\beta),\quad a=\frac{c\sin\alpha}{\sin% \gamma},\quad b=\frac{c\sin\beta}{\sin\gamma}$
2. 2.

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

 $\cos\alpha=\frac{b^{2}\!+\!c^{2}\!-\!a^{2}}{2bc},\quad\cos\beta=\frac{c^{2}\!+% \!a^{2}\!-\!b^{2}}{2ca},\quad\cos\gamma=\frac{a^{2}\!+\!b^{2}\!-\!c^{2}}{2ab}.$
3. 3.

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

 $a^{2}=b^{2}\!+\!c^{2}\!-\!2bc\cos\alpha,\quad\sin\beta=\frac{b\sin\alpha}{a},% \quad\sin\gamma=\frac{c\sin\alpha}{a}$
4. 4.

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

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

Since the SSA criterion alone does not prove congruence, it is not surprising that there may not always be a single solution for $\beta$ 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  $\sin\beta<1$,  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  $\beta_{1}\leq\alpha$,  then there is only one triangle.

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

Title triangle solving TriangleSolving 2013-03-22 15:44:02 2013-03-22 15:44:02 stevecheng (10074) stevecheng (10074) 11 stevecheng (10074) Definition msc 51-00 Congruence