Hansen's_problem Knowpia

In trigonometry, 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 $A, B$ , and two unknown points $P 1, P 2$ . 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 edit

Define the following angles:

{\begin{alignedat}{5}\gamma &=\angle P_{1}AP_{2},&\quad \delta &=\angle P_{1}BP_{2},\\[4pt]\phi &=\angle P_{2}AB,&\quad \psi &=\angle P_{1}BA.\end{alignedat}}

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 equation

\phi +\psi =\beta _{1}+\beta _{2}.

A second equation can be found more laboriously, as follows. The law of sines yields

{\frac {\overline {AB}}{\overline {P_{2}B}}}={\frac {\sin \alpha _{2}}{\sin \phi }},\qquad {\frac {\overline {P_{2}B}}{\overline {P_{1}P_{2}}}}={\frac {\sin \beta _{1}}{\sin \delta }}.

Combining these, we get

{\frac {\overline {AB}}{\overline {P_{1}P_{2}}}}={\frac {\sin \alpha _{2}\sin \beta _{1}}{\sin \phi \sin \delta }}.

Entirely analogous reasoning on the other side yields

{\frac {\overline {AB}}{\overline {P_{1}P_{2}}}}={\frac {\sin \alpha _{1}\sin \beta _{2}}{\sin \psi \sin \gamma }}.

Setting these two equal gives

{\frac {\sin \phi }{\sin \psi }}={\frac {\sin \gamma \sin \alpha _{2}\sin \beta _{1}}{\sin \delta \sin \alpha _{1}\sin \beta _{2}}}=k.

Using a known trigonometric identity this ratio of sines can be expressed as the tangent of an angle difference:

\tan {\tfrac {1}{2}}(\phi -\psi )={\frac {k-1}{k+1}}\tan {\tfrac {1}{2}}(\phi +\psi ).

Where $k={\frac {\sin \phi }{\sin \psi }}.$

This is the second equation we need. Once we solve the two equations for the two unknowns $φ, ψ$ , we can use either of the two expressions above for ${\tfrac {\overline {AB}}{\overline {P_{1}P_{2}}}}$ to find ${\overline {P_{1}P_{2}}}$ since $AB$ is known. We can then find all the other segments using the law of sines.^[1]

Solution algorithm edit

We are given four angles $(α 1, β 1, α 2, β 2)$ and the distance $AB$ . The calculation proceeds as follows:

Calculate ${\begin{aligned}\gamma &=\pi -\alpha _{1}-\beta _{1}-\beta _{2},\\\delta &=\pi -\alpha _{2}-\beta _{1}-\beta _{2}.\end{aligned}}$
Calculate $k={\frac {\sin \gamma \sin \alpha _{2}\sin \beta _{1}}{\sin \delta \sin \alpha _{1}\sin \beta _{2}}}.$
Let $s=\beta _{1}+\beta _{2},\quad d=2\arctan \left({\frac {k-1}{k+1}}\tan {\tfrac {1}{2}}s\right)$ and then $\phi ={\frac {s+d}{2}},\quad \psi ={\frac {s-d}{2}}.$
Calculate ${\overline {P_{1}P_{2}}}={\overline {AB}}\ {\frac {\sin \phi \sin \delta }{\sin \alpha _{2}\sin \beta _{1}}}$ or equivalently ${\overline {P_{1}P_{2}}}={\overline {AB}}\ {\frac {\sin \psi \sin \gamma }{\sin \alpha _{1}\sin \beta _{2}}}.$ If one of these fractions has a denominator close to zero, use the other one.

References edit

^ Udo Hebisch: Ebene und Sphaerische Trigonometrie, Kapitel 1, Beispiel 4 (2005, 2006)[1]

[1] Udo Hebisch: Ebene und Sphaerische Trigonometrie, Kapitel 1, Beispiel 4 (2005, 2006)[1]

[1]

Summary

Solution method overview edit

Solution algorithm edit

See also edit

References edit