![{\displaystyle \operatorname {Sin} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/232dc5b1980bb31e343c75c2145d232d2527a659)
(
XXXVIII)
![{\displaystyle \operatorname {Cos} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3471aa777dfae9578bd565f5947645b23029c8f2)
(
XXXVII)
d’où encore
![{\displaystyle \operatorname {Tang} .{\frac {1}{2}}\left\{{\frac {1}{2}}\varpi -(S-A)\right\}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b0c38f7fee656bbff423eab7af87770e6cf6abf)
(
XXXIX)
§. IV.
Recherches diverses.
Soit
l’arc de grand cercle mené du sommet
au milieu
du côté
il divisera notre triangle en deux autres, pour lesquels on aura (i)
![{\displaystyle {\begin{aligned}&\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .d\operatorname {Cos} .AA'C=\operatorname {Cos} .b-\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .d,\\\\&\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .d\operatorname {Cos} .AA'B=\operatorname {Cos} .c-\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .d\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/868cd28825da0a2493e21823fdaebf1ca1b03414)
en prenant la somme de ces deux équations et observant que
![{\displaystyle \operatorname {Cos} .AA'C+\operatorname {Cos} .AA'B=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/901d0dfdc03a2adaaa77bdcae633cbcf2ad9001e)
il viendra, en transposant,
![{\displaystyle 2\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .d=\operatorname {Cos} .b+\operatorname {Cos} .c=2\operatorname {Cos} .{\frac {1}{2}}(b+c)\operatorname {Cos} .{\frac {1}{2}}(b-c)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4a40eeb8b096bc8688763427774d2fd9eaabc9e)
et, par suite,