Page:Annales de mathématiques pures et appliquées, 1817-1818, Tome 8.djvu/258

248

RECTIFICATION

$\mathrm {AF=AD} =2\operatorname {Sin} .60^{\circ }={\sqrt {3}}\,;$

donc

$\mathrm {CF} ={\sqrt {{\overline {\mathrm {AF} }}^{2}-{\overline {\mathrm {AC} }}^{2}}}={\sqrt {3-1}}={\sqrt {2}}\,;$

si donc $\mathrm {H}$ est l’intersection de $\mathrm {CF}$ et $\mathrm {BD} \,;$ comme $\mathrm {CH} =\operatorname {Sin} .60^{\circ }={\tfrac {1}{2}}{\sqrt {3}}\,;$ on aura

$\mathrm {FH=CF-CH} ={\sqrt {2}}-{\tfrac {1}{2}}{\sqrt {3}}={\frac {2{\sqrt {2}}-{\sqrt {3}}}{2}}.$

Le triangle $\mathrm {BHF}$ donnera ensuite

$\mathrm {BG=BF} ={\sqrt {{\overline {\mathrm {BH} }}^{2}+{\overline {\mathrm {HF} }}^{2}}}={\sqrt {{\frac {1}{4}}+{\frac {\left(2{\sqrt {2}}-{\sqrt {3}}\right)^{2}}{4}}}}\,;$

c’est-à-dire,

$\mathrm {BG} ={\sqrt {3-{\sqrt {6}}}}.$

On aura d’après cela

$Sin.{\tfrac {1}{2}}Arc\mathrm {BG} ={\tfrac {1}{2}}{\sqrt {3-{\sqrt {6}}}},\qquad Sin.{\tfrac {1}{2}}Arc\mathrm {AB} ={\tfrac {1}{2}},$

d’où

$Cos.{\tfrac {1}{2}}Arc\mathrm {BG} ={\tfrac {1}{2}}{\sqrt {1+{\sqrt {6}}}},\qquad Cos.{\tfrac {1}{2}}Arc\mathrm {AB} ={\tfrac {1}{2}}{\sqrt {3}}.$

On aura donc enfin

${\begin{aligned}\mathrm {AG} &=2Sin.{\tfrac {1}{2}}\mathrm {ABG} =2Sin.\left({\tfrac {1}{2}}Arc.\mathrm {AB} +{\tfrac {1}{2}}Arc.\mathrm {BG} \right)\\&=2Sin.{\tfrac {1}{2}}Arc.\mathrm {AB} Cos.{\tfrac {1}{2}}Arc.\mathrm {BG} +2Sin.{\tfrac {1}{2}}Arc.\mathrm {BG} Cos.{\tfrac {1}{2}}Arc.\mathrm {AB} \\\end{aligned}}$

c’est-à-dire, en substituant,