Tangent of Half Angle in Triangle

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Theorem

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Let $s$ denote the semiperimeter of $\triangle ABC$:

$s = \dfrac {a + b + c} 2$

Then:

$\tan \dfrac C 2 = \sqrt {\dfrac {\paren {s - a} \paren {s - b} } {s \paren {s - c} } }$


Proof

\(\ds \tan \dfrac C 2\) \(=\) \(\ds \dfrac {\sin \dfrac C 2} {\cos \dfrac C 2}\)
\(\ds \) \(=\) \(\ds \dfrac {\sqrt {\dfrac {\paren {s - a} \paren {s - b} } {a b} } } {\sqrt {\dfrac {s \paren {s - c} } {a b} } }\) Sine of Half Angle in Triangle, Cosine of Half Angle in Triangle
\(\ds \) \(=\) \(\ds \sqrt {\dfrac {\paren {s - a} \paren {s - b} } {s \paren {s - c} } }\) after simplification

$\blacksquare$


Also presented as

The Tangent of Half Angle in Triangle formula is also seen presented in the following form:

\(\ds \tan \dfrac A 2\) \(=\) \(\ds \dfrac r {s - a}\)
\(\ds \tan \dfrac B 2\) \(=\) \(\ds \dfrac r {s - b}\)
\(\ds \tan \dfrac C 2\) \(=\) \(\ds \dfrac r {s - c}\)

where:

\(\ds s\) \(=\) \(\ds \dfrac {a + b + c} 2\)
\(\ds r\) \(=\) \(\ds \sqrt {\dfrac {\paren {s - a} \paren {s - b} \paren {s - c} } s}\)


Also see


Historical Note

In the days before electronic calculators, Tangent of Half Angle in Triangle was often used in preference to the Cosine Rule because it was more convenient for use with log tables.


Sources

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