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MÉTHODE D’OLBERS.
(taches sur le fac-similé, voir la page de discussion)
Nous avons, dans le triangle
en supposant le plan de projection
en
![{\displaystyle \mathrm {A} _{1}'a_{1}=\rho \sin(\Theta '-\alpha )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76bd84802e4b471ca11f3b15cb8ee3c1fdfd9650)
et, dans le triangle ![{\displaystyle \mathrm {A} _{1}\mathrm {A} _{1}'a_{1},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc44718a8ed1bd42386a8e640140b04abd5c9b62)
![{\displaystyle \mathrm {A} _{1}'a_{1}=\delta \cos b\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ef66700814552e990299bd4194b0fc1e4e96f13)
on en déduit
![{\displaystyle \rho ={\frac {\delta \cos b}{\sin(\Theta '-\alpha )}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/276ef07d1d979b8cd63d276d6c1842181ce7f2ad)
En supposant le plan de projection en
nous obtiendrons, par des
triangles analogues,
![{\displaystyle \rho ''={\frac {\mathrm {N} \delta \cos b''}{\sin(\Theta '-\alpha '')}}=\mathrm {M} \rho .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bead0ecefed1229e93ff9a5e985f64172116db47)
Les triangles
et
donnent
![{\displaystyle {\begin{aligned}\mathrm {C} _{1}o&={\frac {\mathrm {D} _{1}\mathrm {C} _{1}\sin \mathrm {D} _{1}}{\sin(b''-b')}},&c_{1}o&={\frac {d_{1}c_{1}\sin d_{1}}{\sin(b''-b')}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dec0653c53c5e016077c45ed521fa397440c34a5)
on en déduit
![{\displaystyle \mathrm {C} _{1}c_{1}={\frac {1}{\sin(b''-b')}}\left(\mathrm {D} _{1}\mathrm {C} _{1}\sin \mathrm {D} _{1}+d_{1}c_{1}\sin d_{1}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0b0ee62ed392f8e145b7d3ae9a28aaead2d6714)
Des deux triangles
et
on obtient de même
![{\displaystyle \mathrm {A} _{1}a_{1}={\frac {1}{\sin(b'-b)}}\left(\mathrm {A} _{1}\mathrm {D} _{1}\sin \mathrm {D} _{1}+a_{1}d_{1}\sin d_{1}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2103644b920a3bd7f6f37823fe5b337e4b366808)
on a par suite,
![{\displaystyle {\frac {\delta ''}{\delta }}=\mathrm {N} ={\frac {\sin(b'-b)}{\sin(b''-b')}}{\frac {\left(\mathrm {D} _{1}\mathrm {C} _{1}\sin \mathrm {D} _{1}+d_{1}c_{1}\sin d_{1}\right)}{\left(\mathrm {A} _{1}\mathrm {D} _{1}\sin \mathrm {D} _{1}+a_{1}d_{1}\sin d_{1}\right)}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37be8deb56b3e2f32f2c99099b83cf9e3f16b42f)
mais on a évidemment,
et
![{\displaystyle {\frac {d_{1}c_{1}}{a_{1}d_{1}}}={\frac {dc}{ad}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6616416f18dad252d9d83e11f4ba6d6ad1cbc5d7)
et comme on a déjà
![{\displaystyle {\frac {\mathrm {DC} }{\mathrm {AD} }}={\frac {dc}{ad}}={\frac {t''}{t'}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/044ff822b398a18a456b3491d3b979906eda1d6d)
il vient alors,
![{\displaystyle {\frac {\mathrm {D} _{1}\mathrm {C} _{1}}{\mathrm {A} _{1}\mathrm {D} _{1}}}={\frac {d_{1}c_{1}}{a_{1}d_{1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a6a3fea84b05abe277c1f06c96f947cb2518d45)
d’où
![{\displaystyle {\frac {\mathrm {D} _{1}\mathrm {C} _{1}\sin \mathrm {D} _{1}+d_{1}c_{1}\sin d_{1}}{\mathrm {A} _{1}\mathrm {D} _{1}\sin \mathrm {D} _{1}+a_{1}d_{1}\sin d_{1}}}={\frac {t''}{t'}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e30ada24a53cd70de8fe02086308dbbe087ff372)