Page:Annales de mathématiques pures et appliquées, 1811-1812, Tome 2.djvu/395

379

RÉSOLUES.

${\frac {b^{2}}{4}}={\frac {c'^{2}}{4\operatorname {Sin} .^{2}\beta '}}-{\frac {2c'c''}{4\operatorname {Sin} .\beta '\cdot \operatorname {Sin} .\beta ''}}\operatorname {Cos} .(\beta +\gamma )+{\frac {c''^{2}}{4\operatorname {Sin} .^{2}\beta ''}},$

c’est-à-dire,

$b^{2}={\frac {c'^{2}\operatorname {Sin} .^{2}\beta ''-2c'c''\operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''\operatorname {Cos} .(\beta +\gamma )+c''^{2}\operatorname {Sin} .^{2}\beta '}{\operatorname {Sin} .^{2}\beta '\operatorname {Sin} .^{2}\beta ''}}.$

Mais, en désignant par $C$ l’aire du triangle $\mathrm {CC'C''} ,$ on a

${\begin{aligned}c'^{2}\operatorname {Sin} .^{2}\beta ''=&c'^{2}\operatorname {Sin} .\beta ''\operatorname {Sin} .(\beta +\beta ')\\=&c'^{2}\operatorname {Sin} .\beta \operatorname {Cos} .\beta '\operatorname {Sin} .\beta ''+c'^{2}\operatorname {Cos} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''\;;\\\end{aligned}}$

${\begin{aligned}-2c'c''\operatorname {Sin} .\beta '\operatorname {Sin} \beta ''\operatorname {Cos} .(\beta +\gamma )&=-2c'c''\operatorname {Cos} .\beta .\operatorname {Sin} .\beta '.\operatorname {Sin} .\beta ''\operatorname {Cos} .\gamma \\&+2c'c''\operatorname {Sin} .\gamma .\operatorname {Sin} .\beta .\operatorname {Sin} .\beta '.\operatorname {Sin} .\beta ''\\\end{aligned}}$

${\begin{aligned}=-&2c'c''\operatorname {Cos} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''\operatorname {Cos} .\gamma +4C\operatorname {Sin} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''\;;\\c''^{2}\operatorname {Sin} .^{2}\beta '&=c''^{2}\operatorname {Sin} .\beta '\operatorname {Sin} .(\beta +\beta '')\\&=c''^{2}\operatorname {Sin} .\beta \operatorname {Sin} .\beta '\operatorname {Cos} .\beta ''+c''^{2}\operatorname {Cos} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''\;;\\\end{aligned}}$

donc

$c'^{2}\operatorname {Sin} .^{2}\beta ''-2c'c''\operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''\operatorname {Cos} .(\beta +\gamma )+c''^{2}\operatorname {Sin} .^{2}\beta '$

$=c'^{2}\operatorname {Sin} .\beta \operatorname {Cos} .\beta '\operatorname {Sin} .\beta ''+c''\operatorname {Sin} .\beta \operatorname {Sin} .\beta '\operatorname {Cos} .\beta ''$

$+(c'^{2}-2c'c''\operatorname {Cos} .\gamma +c''^{2})\operatorname {Cos} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''+4C\operatorname {Sin} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''$

$=c^{2}\operatorname {Cos} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''+c'^{2}\operatorname {Sin} .\beta \operatorname {Cos} .\beta '\operatorname {Sin} .\beta ''$

$+c''^{2}\operatorname {Sin} .\beta \operatorname {Sin} .\beta '\operatorname {Cos} .\beta ''+4C\operatorname {Sin} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''.$

Posant donc

$\mathrm {M^{2}} =\left\{{\begin{aligned}&c^{2}\operatorname {Cos} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''\\+&c'^{2}\operatorname {Sin} .\beta \operatorname {Cos} .\beta '\operatorname {Sin} .\beta ''+4C\operatorname {Sin} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta '',\\+&c''^{2}\operatorname {Sin} .\beta \operatorname {Sin} .\beta '\operatorname {Cos} .\beta ''\\\end{aligned}}\right.$

il viendra

$b={\frac {\mathrm {M} }{\operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''}},\qquad b'={\frac {\mathrm {M} }{\operatorname {Sin} .\beta ''\operatorname {Sin} .\beta }},\qquad b''={\frac {\mathrm {M} }{\operatorname {Sin} .\beta \operatorname {Sin} .\beta '}},$

Si l’on désigne par $B$ l’aire du triangle $\mathrm {BB'B'',}$ on aura

$B={\frac {bb'\operatorname {Sin} .\beta ''}{2}}={\frac {\mathrm {M} ^{2}}{2\operatorname {Sin} .\beta \operatorname {Sin} .\beta '\operatorname {Sin} .\beta ''}},$

et par conséquent