Page:Mémoires de l’Académie des sciences, Tome 10.djvu/758

${\begin{alignedat}{2}\operatorname {l} .\sin .\mathrm {Z} =&9.8140223\\\operatorname {l} .\cos .\mathrm {Z} =&9.8799610-&\operatorname {l} .\sin .2\mathrm {Z} =&9.6280446-\\\operatorname {l} .\cot .p=&{\underline {1.9566915}}&\operatorname {l} .\sin .\lambda '=&{\underline {9.8395782}}\\&1.6506748-&&9.4676228-\\=&-44{,}737810&=&0{,}293509\\&-0{,}293509\\\mathrm {M} =&-45{,}031319&\log .\mathrm {M} =-1.6535147-&\end{alignedat}}$

Évaluant ensuite

$\cos .\lambda _{0}=\cos .\lambda '\sin .\mathrm {Z} ,\qquad \cos .\sigma _{0}'={\frac {\sin .\lambda '}{\sin .\lambda _{0}}},\qquad \cos .\sigma _{0}''={\frac {\sin .\lambda ''}{\sin .\lambda _{0}}},$

on trouvera

$\lambda _{0}=68^{\text{g}}{,}7816''{,}95\,;\qquad \sigma '_{0}=42^{\text{g}}{,}6875''{,}92,\qquad \sigma ''_{0}=43^{\text{g}}{,}4746''{,}31.$

Au moyen de ces valeurs la série (F) donnera, à cause de $\sigma ''_{0}-\sigma '_{0}=7870''{,}39$ centésimales,

$\left.{\begin{alignedat}{2}\log .(\sigma ''_{0}-\sigma '_{0})=&3.8959963&\ldots \ldots \ldots \ldots &3.89600\\\log .{\frac {1}{2}}\varepsilon =&7.5126590&\qquad \qquad \qquad \log {\frac {1}{16}}\varepsilon ^{2}=&4.42326-\\\operatorname {l} .\cos .\lambda _{0}=&{\underline {9.6729818}}&\ldots \ldots \ldots \ldots &9.672980\end{alignedat}}\right\}(\Sigma )$

${\begin{alignedat}{2}&1.0816371\qquad \qquad \qquad &\operatorname {l} .\sin .^{2}\lambda _{0}=&{\underline {9.89109}}\\&=12'',068&&7.88333-\\&&=-&0''{,}0076\end{alignedat}}$

${\begin{alignedat}{2}\left(\Sigma \right)=7.99224-\quad &&&{\underline {-0'',0589}}\\\log .6{\underline {=0.77815}}\qquad &\\&&&-0{,}0665\\8.77039-=&0''{,}0589&&+12{,}068\\\log .{\frac {1}{16}}\varepsilon ^{2}\cos .\lambda _{0}\sin .^{2}\lambda _{0}=&3.98733&2^{\text{e}}{\text{ terme}}=&+12{,}002\\c.\log .\sin .1''=&5.80388\\\log .{\frac {1}{2}}=&{\underline {9.69897}}\\&9.49018&\ldots \ldots \ldots \ldots &9.49018\\\log .2\sigma '_{0}=&{\underline {9.98843}}\,;&\operatorname {l} .\sin .2.\sigma '_{0}=&{\underline {9.99081-}}\\&9.47861+&&9.48099-\\=+&0''{,}30103&=-&0''{,}30269\\&&{\underline {+}}&{\underline {0\ \ {,}30103}}\\&&3^{\text{e}}{\text{ terme}}={\underline {-}}&{\underline {0\ \ {,}00166}}\end{alignedat}}$