Page:Annales de mathématiques pures et appliquées, 1810-1811, Tome 1.djvu/301

291

RÉSOLUES.

d’où, à cause de $\mathrm {A+B+C=360^{\circ }}$, on conclura

$\mathrm {A=B=C,}$

comme ci-dessus.

2.^o Pour le quadrilatère, on trouvera, en faisant absolument le même calcul,

$a=\mathrm {\tfrac {\lambda \operatorname {Cos} .B}{\operatorname {Sin} .(A+B)}} ,\quad b=\mathrm {\tfrac {\lambda \operatorname {Cos} .C}{\operatorname {Sin} .(B+C)}} ,\quad c=\mathrm {\tfrac {\lambda \operatorname {Cos} .D}{\operatorname {Sin} .(C+D)}} ,\quad d=\mathrm {\tfrac {\lambda \operatorname {Cos} .A}{\operatorname {Sin} .(D+A)}} \ ;$

$a=\mathrm {\tfrac {\lambda \operatorname {Cos} .C}{\operatorname {Sin} .(C+D)}} ,\quad b=\mathrm {\tfrac {\lambda \operatorname {Cos} .D}{\operatorname {Sin} .(D+A)}} ,\quad c=\mathrm {\tfrac {\lambda \operatorname {Cos} .A}{\operatorname {Sin} .(A+B)}} ,\quad d=\mathrm {\tfrac {\lambda \operatorname {Cos} .B}{\operatorname {Sin} .(B+C)}} \ ;$

d’où on conclura :

$\operatorname {Cos} .\mathrm {B} \operatorname {Sin} .\mathrm {(C+D)} =\operatorname {Cos} .\mathrm {C} \operatorname {Sin} .\mathrm {(A+B)} ,\quad \operatorname {Cos} .\mathrm {C} \operatorname {Sin} .\mathrm {(D+A)} =\operatorname {Cos} .\mathrm {D} \operatorname {Sin} .\mathrm {(B+C)} ,$

$\operatorname {Cos} .\mathrm {D} \operatorname {Sin} .\mathrm {(A+B)} =\operatorname {Cos} .\mathrm {A} \operatorname {Sin} .\mathrm {(C+D)} ,\quad \operatorname {Cos} .\mathrm {A} \operatorname {Sin} .\mathrm {(B+C)} =\operatorname {Cos} .\mathrm {B} \operatorname {Sin} .\mathrm {(D+A)} .$

La condition

$\mathrm {A+B+C+D=360^{\circ }}$

donne d’ailleurs

$\operatorname {Sin} .\mathrm {(C+D)} =-\operatorname {Sin} .\mathrm {(A+B)} ,\qquad \operatorname {Sin} .\mathrm {(D+A)} =-\operatorname {Sin} .\mathrm {(B+C)} .$

substituant donc, il viendra

$\operatorname {Cos} .\mathrm {B} =-\operatorname {Cos} .\mathrm {C} ,\ \operatorname {Cos} .\mathrm {C} =-\operatorname {Cos} .\mathrm {D} ,\ \operatorname {Cos} .\mathrm {D} =-\operatorname {Cos} .\mathrm {A} ,\ \operatorname {Cos} .\mathrm {A} =-\operatorname {Cos} .\mathrm {B} ~;$

et encore

$\mathrm {A=C,\qquad B=D\ ;}$

mêmes relations que ci-dessus.