![{\displaystyle \operatorname {Sin} .X={\frac {2r{\sqrt {a^{2}-r^{2}}}}{a^{2}}},\ \ \operatorname {Sin} .Y={\frac {2r{\sqrt {b^{2}-r^{2}}}}{b^{2}}},\ \ \operatorname {Sin} .Z={\frac {2r{\sqrt {c^{2}-r^{2}}}}{c^{2}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a0735de3968f03fd64a33c4ff3444d02c050ee2)
Si l’on représente par
les trois côtés de ce même triangle, on aura
![{\displaystyle {\begin{aligned}&x=b\operatorname {Sin} .\beta +c\operatorname {Sin} .\gamma ,\\&y=c\operatorname {Sin} .\gamma +a\operatorname {Sin} .\alpha ,\\&z=a\operatorname {Sin} .\alpha +b\operatorname {Sin} .\beta \,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd80f2e66949e28e85da6d604866e5ab20ad3efd)
c’est-à-dire,
![{\displaystyle {\begin{aligned}&x={\sqrt {b^{2}-r^{2}}}+{\sqrt {c^{2}-r^{2}}},\\&y={\sqrt {c^{2}-r^{2}}}+{\sqrt {a^{2}-r^{2}}},\\&z={\sqrt {a^{2}-r^{2}}}+{\sqrt {b^{2}-r^{2}}}\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e481fb17d6ff31bdec3370da5991493e3bf72fa6)
/o^2-r^2+tc^2-r mais des valeurs des sinus des angles, données ci-dessus, on tire
![{\displaystyle {\sqrt {a^{2}-r^{2}}}={\frac {a^{2}\operatorname {Sin} .X}{2r}},\ \ {\sqrt {b^{2}-r^{2}}}={\frac {b^{2}\operatorname {Sin} .Y}{2r}},\ \ {\sqrt {c^{2}-r^{2}}}={\frac {c^{2}\operatorname {Sin} .Z}{2r}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/092f553093d6b0f4f92b00363c4b5aa99303884b)
ce qui donnera, en substituant,
![{\displaystyle {\begin{aligned}&x={\frac {b^{2}\operatorname {Sin} .Y+c^{2}\operatorname {Sin} .Z}{2r}},\\\\&y={\frac {c^{2}\operatorname {Sin} .Z+a^{2}\operatorname {Sin} .X}{2r}},\\\\&z={\frac {a^{2}\operatorname {Sin} .X+b^{2}\operatorname {Sin} .Y}{2r}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39c0bd1dc53c81f50619f45c2da445a9e47e0db6)
Toutes ces valeurs étant fonction de
, qui est donné par