24. En second lieu, on aura
![{\displaystyle {\begin{aligned}&{\frac {c+b-a}{2}}\\&\quad =\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\sin a+{\frac {1}{2}}\left(\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\right)^{2}\sin 2a\\&\qquad \qquad \qquad \qquad \qquad +{\frac {1}{3}}\left(\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\right)^{3}\sin 3a+\ldots \\&\quad =\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\sin(b+c)-{\frac {1}{2}}\left(\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\right)^{2}\sin 2(b+c)\\&\qquad \qquad \qquad \qquad \qquad +{\frac {1}{3}}\left(\operatorname {tang} {\frac {\beta }{2}}\operatorname {tang} {\frac {\gamma }{2}}\right)^{3}\sin 3(b+c)-\ldots .\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f3d870eca7d4226e8fcc58ebcc245138524bc4b)
25. Enfin la cinquième équation du no 20, étant comparée à celle du no 10, donnera
![{\displaystyle \varphi ={\frac {\beta -\gamma }{2}},\quad \omega ={\frac {\beta +\gamma }{2}},\quad x={\frac {b+c}{2}},\quad y={\frac {b-c}{2}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea9a0650f54e59d4e34114d244b142d39af28fe9)
donc
![{\displaystyle \theta =-{\frac {\sin \gamma }{\sin \beta }},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26986756d51f0d429686ee209dc69d0e58948020)
et par conséquent
![{\displaystyle {\begin{aligned}c=&{\frac {\sin \gamma }{\sin \beta }}\sin(b-c)+{\frac {1}{2}}\left({\frac {\sin \gamma }{\sin \beta }}\right)^{2}\sin 2(b-c)+{\frac {1}{3}}\left({\frac {\sin \gamma }{\sin \beta }}\right)^{3}\sin 3(b-c)+\ldots \\=&{\frac {\sin \gamma }{\sin \beta }}\sin(b+c)-{\frac {1}{2}}\left({\frac {\sin \gamma }{\sin \beta }}\right)^{2}\sin 2(b+c)+{\frac {1}{3}}\left({\frac {\sin \gamma }{\sin \beta }}\right)^{3}\sin 3(b+c)-\ldots .\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20b043fcbd779358115e182d05f7684a5546fb43)
On peut aussi comparer d’une autre manière la même équation avec celle du numéro cité, en faisant
![{\displaystyle \varphi ={\frac {b+c}{2}},\quad \omega ={\frac {b-c}{2}},\quad x={\frac {\beta -\gamma }{2}},\quad y={\frac {\beta +\gamma }{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e141815cb816291c5b4de24bfcaf9569f9bf0c94)
ce qui donnera
![{\displaystyle \theta ={\frac {\sin c}{\sin b}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73fd703f57b2d95c0964717ddc9e9b5fa386afd6)
et de là par le no 7
![{\displaystyle {\begin{aligned}\gamma =&{\frac {\sin c}{\sin b}}\sin(\beta +\gamma )-{\frac {1}{2}}\left({\frac {\sin c}{\sin b}}\right)^{2}\sin 2(\beta +\gamma )+{\frac {1}{3}}\left({\frac {\sin c}{\sin b}}\right)^{3}\sin 3(\beta +\gamma )-\ldots \\=&{\frac {\sin c}{\sin b}}\sin(\beta -\gamma )+{\frac {1}{2}}\left({\frac {\sin c}{\sin b}}\right)^{2}\sin 2(\beta -\gamma )+{\frac {1}{3}}\left({\frac {\sin c}{\sin b}}\right)^{3}\sin 3(\beta -\gamma )+\ldots .\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/111d96a515aa9bde7a316f6fa74cd3c51dca8e3d)