Page:Annales de mathématiques pures et appliquées, 1811-1812, Tome 2.djvu/290

278

QUESTIONS

Soit $\mathrm {AA'A''A'''}$ un quadrilatère auquel on doit inscrire un autre quadrilatère $\mathrm {XX'X''X''',}$ dont les côtés soient respectivement parallèles à des droites données.

Que les angles du premier quadrilatère soient désignés par $\mathrm {A,A',A'',A'''}$ et soient faits

${\begin{array}{llll}\operatorname {Ang} .\mathrm {AXX'''} &=\alpha ,&\operatorname {Ang} .\mathrm {A'XX'} &=\beta ,\\\operatorname {Ang} .\mathrm {A'X'X} &=\alpha ',&\operatorname {Ang} .\mathrm {A''X'X''} &=\beta ',\\\operatorname {Ang} .\mathrm {A''X''X'} &=\alpha '',&\operatorname {Ang} .\mathrm {A'''X''X'''} &=\beta '',\\\operatorname {Ang} .\mathrm {A'''X'''X''} &=\alpha ''',&\operatorname {Ang} .\mathrm {AX'''X} &=\beta '''\,;\\\end{array}}$

soit enfin $\mathrm {XA'} =x,$ on aura

$\mathrm {A'X'} =x\cdot {\tfrac {\operatorname {Sin} .\beta }{\operatorname {Sin} .\alpha '}},\quad \mathrm {A''X'} =\mathrm {A'A''} -x.{\tfrac {\operatorname {Sin} .\beta }{\operatorname {Sin} .\alpha '}},$

$\mathrm {A''X''} =\mathrm {A'A''} .{\tfrac {\operatorname {Sin} .\beta '}{\operatorname {Sin} .\alpha ''}}-x.{\tfrac {\operatorname {Sin} .\beta .\operatorname {Sin} .\beta '}{\operatorname {Sin} .\alpha '.\operatorname {Sin} .\alpha ''}},\ \mathrm {A'''X''=A''A'''-A'A''} .{\tfrac {\operatorname {Sin} .\beta '}{\operatorname {Sin} .\alpha ''}}+x.{\tfrac {\operatorname {Sin} .\beta .\operatorname {Sin} .\beta '}{\operatorname {Sin} .\alpha '.\operatorname {Sin} .\alpha ''}},$

${\begin{array}{ll}\mathrm {A'''X'''} &=\mathrm {A''A'''} .{\tfrac {\operatorname {Sin} .\beta ''}{\operatorname {Sin} .\alpha '''}}-\mathrm {A'A''} .{\tfrac {\operatorname {Sin} .\beta '.\operatorname {Sin} .\beta ''}{\operatorname {Sin} .\alpha ''.\operatorname {Sin} .\alpha '''}}+x.{\tfrac {\operatorname {Sin} .\beta .\operatorname {Sin} .\beta '.\operatorname {Sin} .\beta ''}{\operatorname {Sin} .\alpha '.\operatorname {Sin} .\alpha ''.\operatorname {Sin} .\alpha '''}},\\\\\mathrm {AX'''} &=\mathrm {A'''A-A''A'''} .{\tfrac {\operatorname {Sin} .\beta ''}{\operatorname {Sin} .\alpha '''}}+\mathrm {A'A''} .{\tfrac {\operatorname {Sin} .\beta '.\operatorname {Sin} .\beta ''}{\operatorname {Sin} .\alpha ''.\operatorname {Sin} .\alpha '''}}-x\cdot {\tfrac {\operatorname {Sin} .\beta .\operatorname {Sin} .\beta '.\operatorname {Sin} .\beta ''}{\operatorname {Sin} .\alpha '.\operatorname {Sin} .\alpha ''.\operatorname {Sin} .\alpha '''}},\\\\\mathrm {AX} &=\mathrm {A'''A} .{\tfrac {\operatorname {Sin} .\beta '''}{\operatorname {Sin} .\alpha }}-\mathrm {A''A'''} .{\tfrac {\operatorname {Sin} .\beta ''.\operatorname {Sin} .\beta '''}{\operatorname {Sin} .\alpha .\operatorname {Sin} .\alpha ''}}+\mathrm {A'A''} \cdot {\tfrac {\operatorname {Sin} .\beta '.\operatorname {Sin} .\beta ''.\operatorname {Sin} .\beta '''}{\operatorname {Sin} .\alpha .\operatorname {Sin} .\alpha '''.\operatorname {Sin} .\alpha ''}},\\\end{array}}$

$-x\cdot {\tfrac {\operatorname {Sin} .\beta .\operatorname {Sin} .\beta '.\operatorname {Sin} .\beta ''.\operatorname {Sin} .\beta '''}{\operatorname {Sin} .\alpha .\operatorname {Sin} .\alpha '''.\operatorname {Sin} .\alpha ''\operatorname {Sin} .\alpha '}}=\mathrm {AA'} -x.$

Cette dernière équation donne

$\left\{1-{\tfrac {\operatorname {Sin} .\beta .\operatorname {Sin} .\beta '.\operatorname {Sin} .\beta ''.\operatorname {Sin} .\beta '''}{\operatorname {Sin} .\alpha .\operatorname {Sin} .\alpha '.\operatorname {Sin} .\alpha ''\operatorname {Sin} .\alpha '''}}\right\}x=$

$\mathrm {AA'-A'A''} .{\tfrac {\operatorname {Sin} .\beta '.\operatorname {Sin} .\beta ''.\operatorname {Sin} .\beta '''}{\operatorname {Sin} .\alpha .\operatorname {Sin} .\alpha '''.\operatorname {Sin} .\alpha ''}}+\mathrm {A''A'''} .{\tfrac {\operatorname {Sin} .\beta ''.\operatorname {Sin} .\beta '''}{\operatorname {Sin} .\alpha .\operatorname {Sin} .\alpha ''}}-\mathrm {A'''A} .{\tfrac {\operatorname {Sin} .\beta '''}{\operatorname {Sin} .\alpha }}\,;$

d’où on tire