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Hansen's problem : ウィキペディア英語版
Hansen's problem

Hansen's problem is a problem in planar surveying, named after the astronomer Peter Andreas Hansen (1795–1874), who worked on the geodetic survey of Denmark. There are two known points ''A'' and ''B'', and two unknown points ''P''1 and ''P''2. From ''P''1 and ''P''2 an observer measures the angles made by the lines of sight to each of the other three points. The problem is to find the positions of ''P''1 and ''P''2. See figure; the angles measured are (''α''1, ''β''1, ''α''2, ''β''2).
Since it involves observations of angles made at unknown points, the problem is an example of resection (as opposed to intersection).
== Solution method overview==

Define the following angles:
''γ'' = ''P''1''AP''2, ''δ'' = ''P''1''BP''2, ''φ'' = ''P''2''AB'', ''ψ'' = ''P''1''BA''.
As a first step we will solve for ''φ'' and ''ψ''.
The sum of these two unknown angles is equal to the sum of ''β''1 and ''β''2, yielding the following equation:
:\phi+\psi=\beta_1+\beta_2
A second equation can be found more laboriously, as follows. The law of sines yields
:\frac=\frac and
:\frac=\frac
combining these together we get
:\frac=\frac
An entirely analogous reasoning on the other side yields
:\frac=\frac
Setting these two equal gives
:\frac=\frac = k
Using a known trigonometric identity this ratio of sines can be expressed as the tangent of an angle difference:
:\tan \frac=\frac\tan\frac
This is the second equation we need. Once we solve the two equations for the two unknowns \phi and \psi, we can use either of the two expressions above for \frac to find ''P''1''P''2 since ''AB'' is known. We can then find all the other segments using the law of sines.〔Udo Hebisch: Ebene und Sphaerische Trigonometrie, Kapitel 1, Beispiel 4 (2005, 2006)()〕

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
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