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Law of Tangents

Let a Triangle have sides of lengths $a$, $b$, and $c$ and let the Angles opposite these sides by $A$, $B$, and $C$. The law of tangents states

\begin{displaymath}
{a-b\over a+b} = {\tan[{\textstyle{1\over 2}}(A-B)]\over\tan[{\textstyle{1\over 2}}(A+B)]}.
\end{displaymath}

An analogous result for oblique Spherical Triangles states that

\begin{displaymath}
{\tan[{\textstyle{1\over 2}}(a-b)]\over\tan[{\textstyle{1\ov...
...style{1\over 2}}(A-B)]\over\tan[{\textstyle{1\over 2}}(A+B)]}.
\end{displaymath}

See also Law of Cosines, Law of Sines


References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 79, 1972.

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp. 145 and 149, 1987.




© 1996-9 Eric W. Weisstein
1999-05-26