On parvient à une équation fort simple en procédant comme il suit :
Soit
, (fig. 2.) l’angle donné, soit
le point donné et soit enfin
la
droite cherchée. Soit mené
soient faits ![{\displaystyle \operatorname {Ang} .\mathrm {PCA} =\alpha ,\ \operatorname {Ang} .\mathrm {PCB} =\beta ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05b863abae945a238f9133e24ffeb03412f96b17)
on aura
donc
![{\displaystyle \left.{\begin{aligned}\operatorname {Ang} .\mathrm {CXP} &=\varpi -(\theta +\alpha ),\\\operatorname {Ang} .\mathrm {CYP} &=(\theta -\beta ),\\\end{aligned}}\right\}{\text{d'où}}\left\{{\begin{aligned}\operatorname {Sin} .\mathrm {CXP} &=\operatorname {Sin} .(\theta +\alpha ),\\\operatorname {Sin} .\mathrm {CYP} &=\operatorname {Sin} .(\theta -\beta ),\\\end{aligned}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b93069b72ae3948f608f6eb1b0597a3611efe581)
donc
![{\displaystyle \mathrm {PY} =\mathrm {K} \cdot {\frac {\operatorname {Sin} .\beta }{\operatorname {Sin} .(\theta -\beta )}},\quad \mathrm {PX} =\mathrm {K} \cdot {\frac {\operatorname {Sin} .\alpha }{\operatorname {Sin} .(\theta +\alpha )}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12d352bd1be380e9b3197b32160c807acc1a397c)
et par conséquent
![{\displaystyle \mathrm {XY=PY+PX=K} \left\{{\frac {\operatorname {Sin} .\beta }{\operatorname {Sin} .(\theta -\beta )}}+{\frac {\operatorname {Sin} .\alpha }{\operatorname {Sin} .(\theta +\alpha )}}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/768b3c297f875885e2c646044b3fafff45c9b88b)
![{\displaystyle =\mathrm {K} \operatorname {Sin} .(\alpha +\beta )\cdot {\frac {\operatorname {Sin} .\theta }{\operatorname {Sin} .(\theta +\alpha )\operatorname {Sin} .(\theta -\beta )}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5438626b3e374db8a7a28af37c8616a7af36acf)
Il faudra donc, pour avoir la valeur de
qui convient au minimum, égaler à zéro
la différentielle de
![{\displaystyle {\frac {\operatorname {Sin} .\theta }{\operatorname {Sin} .(\theta +\alpha )\operatorname {Sin} .(\theta -\beta )}}~;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d5a331138f0c353105e19a9a66b53bf2783ce5a)
ce qui donnera
![{\displaystyle \operatorname {Sin} .(\theta +\alpha )\operatorname {Sin} .(\theta -\beta )\operatorname {Cos} .\theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf958aaa5fbbd676878d9b70cc42d934244b8e90)
![{\displaystyle -\left\{\operatorname {Sin} .(\theta +\alpha )\operatorname {Cos} .(\theta -\beta )+\operatorname {Sin} .(\theta -\beta )\operatorname {Cos} .(\theta +\alpha )\right\}\operatorname {Sin} .\theta =0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43b57a17bd0025ea5f34f78fc400d0ffed1351d4)
En divisant cette équation par
elle devient
![{\displaystyle \operatorname {Cot} .\theta =\operatorname {Cot} .(\theta +\alpha )\operatorname {Cot} .(\theta -\beta )~;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd1e822fdf0860eefffa0c3385f62f516264ed69)
équation équivalente à celle-ci
![{\displaystyle \operatorname {Cot} .\theta ={\frac {\operatorname {Cot} .\alpha \operatorname {Cot} .\theta -1}{\operatorname {Cot} .\alpha +\operatorname {Cot} .\theta }}+{\frac {\operatorname {Cot} .\beta \operatorname {Cot} .\theta +1}{\operatorname {Cot} .\beta -\operatorname {Cot} .\theta }},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67a28ff6f35fab57f9a4efb8dc99f5d6ea46ccd7)
laquelle devient, en chassant les dénominateurs et réduisant,
![{\displaystyle \operatorname {Cot} .^{3}\theta +(2+\operatorname {Cot} .\alpha \operatorname {Cot} .\beta )\operatorname {Cot} .\theta +(\operatorname {Cot} .\alpha -\operatorname {Cot} .\beta )=0~;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b5c1d047f3c891862287efa26cd6565d7d64cc1)
équation du troisième degré, sans second terme.
(Note des éditeurs.)