triangle solvingTriangleSolving2006-03-05 15:22:562007-03-14 20:11:57Definitiontriangle% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}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:
\begin{enumerate}
\item \textbf{ASA.}\, Known two angles and one side, e.g. $\alpha$, $\beta$, $c$.\, Other parts:
$$\gamma = 180\sp\circ\!-\!(\alpha\!+\!\beta), \quad
a = \frac{c\sin\alpha}{\sin\gamma},\quad b = \frac{c\sin\beta}{\sin\gamma}$$
\begin{figure}[!htb]
\begin{center}
\includegraphics{triangle.1.eps}
\end{center}
\caption{ASA (angle-side-angle)}
\end{figure}
\item \textbf{SSS.}\, Known all sides $a$, $b$, $c$.\, The angles are obtained from
$$\cos\alpha = \frac{b^2\!+\!c^2\!-\!a^2}{2bc},
\quad\cos\beta = \frac{c^2\!+\!a^2\!-\!b^2}{2ca},
\quad\cos\gamma = \frac{a^2\!+\!b^2\!-\!c^2}{2ab}.$$
\begin{figure}[!htb]
\begin{center}
\includegraphics{triangle.2.eps}
\end{center}
\caption{SSS (side-side-side)}
\end{figure}
\item \textbf{SAS.}\, Known two sides and the angle between them, e.g. $b$, $c$, $\alpha$.\, Other parts from
$$a^2 = b^2\!+\!c^2\!-\!2bc\cos\alpha, \quad \sin\beta = \frac{b\sin\alpha}{a},
\quad \sin\gamma = \frac{c\sin\alpha}{a}$$
\begin{figure}[!htb]
\begin{center}
\includegraphics{triangle.3.eps}
\end{center}
\caption{SAS (side-angle-side)}
\end{figure}
\item \textbf{SSA.}\, Known two sides and the angle \PMlinkescapetext{opposite} of one of them, e.g. $a$, $b$, $\alpha$.\, Other parts are gotten from
$$\sin\beta = \frac{b\sin\alpha}{a},
\quad \gamma = 180\sp\circ\!-\!(\alpha\!+\!\beta),
\quad c = \frac{a\sin\gamma}{\sin\alpha}.$$
\begin{figure}[!htb]
\begin{center}
\includegraphics{triangle.4.eps}
\end{center}
\caption{SSA (side-side-angle)}
\end{figure}
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\, $\beta_2 = 180\sp\circ-\beta_1$.\, If in this case\, $\beta_1 > \alpha$,\, then there are two different triangles as \PMlinkescapetext{solutions}, but if\,
$\beta_1 \le \alpha$,\, then there is only one triangle.
\end{enumerate}
\begin{itemize}
\item
\PMlinkexternal{MetaPost source code for the above diagrams}{http://svn.gold-saucer.org/repos/PlanetMath/TriangleSolving/triangle.mp}
\end{itemize}