Tangent of Half Side for Spherical Triangles
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Theorem
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
- $\tan \dfrac a 2 = \sqrt {\dfrac {-\cos S \, \map \cos {S - A} } {\map \cos {S - B} \, \map \cos {S - C} } }$
where $S = \dfrac {a + b + c} 2$.
Proof
| \(\ds \tan \dfrac a 2\) | \(=\) | \(\ds \dfrac {\sqrt {\dfrac {-\cos S \, \map \cos {S - A} } {\sin B \sin C} } } {\sqrt {\dfrac {\map \cos {S - B} \, \map \cos {S - C} } {\sin B \sin C} } }\) | Sine of Half Angle for Spherical Triangles, Cosine of Half Angle for Spherical Triangles | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sqrt {\dfrac {-\cos S \, \map \cos {S - A} } {\map \cos {S - B} \, \map \cos {S - C} } }\) | simplification |
Hence the result.
$\blacksquare$
Also presented as
The Tangent of Half Side for Spherical Triangles formula is also seen presented in the following form:
| \(\ds \tan \dfrac a 2\) | \(=\) | \(\ds R \map \cos {S - A}\) | ||||||||||||
| \(\ds \tan \dfrac b 2\) | \(=\) | \(\ds R \map \cos {S - B}\) | ||||||||||||
| \(\ds \tan \dfrac c 2\) | \(=\) | \(\ds R \map \cos {S - C}\) |
where:
| \(\ds S\) | \(=\) | \(\ds \dfrac {A + B + C} 2\) | ||||||||||||
| \(\ds R\) | \(=\) | \(\ds \sqrt {\dfrac {-\cos S} {\map \cos {S - A} \map \cos {S - B} \map \cos {S - C} } }\) |
Also see
- The other Half Side Formulas for Spherical Triangles: