Page:Optics.djvu/55

 {| The angular distances between the object and the images, that is, the angles $OO′=2OA=2a$, $OO″=2O′B−O′A=O′B+OB=2AB=2c$, &c. may be calculated by means of their tangents; thus, if $OO‴=2A′O″−OO″=AO″+AO=2OO″+2AO=2c+2a$ be perpendicular to $OO_{′}=2OB=b$, Thus supposing the distance $OO_{″}=2AO_{′}−OO_{′}=AO_{′}+AO=2AB=2c$ is $OO_{‴}=2BO_{″}−OO_{″}=BO_{″}+BO=OO_{″}+2BO=2c+2b$ inches,
 * }
 * }
 * }
 * }
 * }
 * }
 * }
 * }
 * }
 * }
 * }
 * }
 * }
 * }
 * }
 * }