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AAS Theorem

\begin{figure}\begin{center}\BoxedEPSF{AASTheorem.epsf}\end{center}\end{figure}

Specifying two angles $A$ and $B$ and a side $a$ uniquely determines a Triangle with Area

\begin{displaymath}
K = {a^2\sin B\sin C\over 2\sin A} = {a^2\sin B\sin(\pi-A-B)\over 2\sin A}.
\end{displaymath} (1)

The third angle is given by
\begin{displaymath}
C=\pi-A-B,
\end{displaymath} (2)

since the sum of angles of a Triangle is 180° ($\pi$ Radians). Solving the Law of Sines
\begin{displaymath}
{a\over\sin A}={b\over\sin B}
\end{displaymath} (3)

for $b$ gives
\begin{displaymath}
b=a{\sin B\over\sin A}.
\end{displaymath} (4)

Finally,
$\displaystyle c$ $\textstyle =$ $\displaystyle b\cos A+a\cos B=a(\sin B\cot A+\cos B)$ (5)
  $\textstyle =$ $\displaystyle a\sin B(\cot A+\cot B).$ (6)

See also AAA Theorem, ASA Theorem, ASS Theorem, SAS Theorem, SSS Theorem, Triangle




© 1996-9 Eric W. Weisstein
1999-05-25