Tangent of Half Angle in Triangle
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 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, was often used in preference to the Cosine Rule because it was more convenient for use with log tables.
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(43)$