4.6 Law of tangents

  Topography notes

 

Triangle_with_notations_2

 

Also called Nepero’s law, but it was described in the 13th century by Persian mathematician Nasir al-Din al-Tusi (1201–74), who also presented the law of sines for plane triangles in his five-volume work Treatise on the Quadrilateral.

\dfrac{a-b}{a+b} = \dfrac{\tan[\frac{1}{2}(\alpha-\beta)]}{\tan[\frac{1}{2}(\alpha+\beta)]} \dfrac{b+a}{b-a} = \dfrac{\tan[\frac{1}{2}(\beta + \alpha)]}{\tan[\frac{1}{2}(\beta - \alpha)]}
 \dfrac{a+c}{a-c} = \dfrac{\tan[\frac{1}{2}(\alpha + \gamma)]}{\tan[\frac{1}{2}(\alpha - \gamma)]}  \dfrac{c+a}{c-a} = \dfrac{\tan[\frac{1}{2}(\gamma + \alpha)]}{\tan[\frac{1}{2}(\gamma - \alpha)]}
 \dfrac{b+c}{b-c} = \dfrac{\tan[\frac{1}{2}(\beta + \gamma)]}{\tan[\frac{1}{2}(\beta - \gamma)]} \dfrac{c+b}{c-b} = \dfrac{\tan[\frac{1}{2}(\gamma + \beta)]}{\tan[\frac{1}{2}(\gamma - \beta)]}

Proof:

To prove the Law of tangents we follow these steps:

  1. Consider law of sines:
    \dfrac{a}{\sin{\alpha}}=\dfrac{b}{\sin{\beta}}~~\Rightarrow ~~ \dfrac{a}{b} = \dfrac{\sin{\alpha}}{\sin{\beta}}
  2. sum b and -b:
    \dfrac{a+b}{a-b}={\sin{\alpha} + \sin{\beta}}{\sin{\alpha} - \sin{\beta}}
  3. Use Prosthaphaeresis formulas:
    \dfrac{a+b}{a-b}=\dfrac{2 \sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2})}}{2\sin{(\frac{\alpha - \beta}{2})}\cos{(\frac{\alpha + \beta}{2})}}
  4. Perform…
    \dfrac{a+b}{a-b}=\tan{(\frac{\alpha + \beta}{2})}\cot{(\frac{\alpha - \beta}{2})}
  5. And finally: \dfrac{a+b}{a-b}=\dfrac{\tan{(\frac{\alpha + \beta}{2})}}{\tan{(\frac{\alpha - \beta}{2})}}

Use the same steps for other formulas.


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