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 P₁ and P₂. From P₁ and P₂ 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 P₁ and P₂. See figure; the angles measured are (α₁, β₁, α₂, β₂).

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 = P₁AP₂, δ = P₁BP₂, φ = P₂AB, ψ = P₁BA. As a first step we will solve for φ and ψ. The sum of these two unknown angles is equal to the sum of β₁ and β₂, yielding the equation

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

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

${\displaystyle {\frac {AB}{P_{2}B}}={\frac {\sin \alpha _{2}}{\sin \phi }}}$ and

${\displaystyle {\frac {P_{2}B}{P_{1}P_{2}}}={\frac {\sin \beta _{1}}{\sin \delta }}.}$

Combining these, we get

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

Entirely analogous reasoning on the other side yields

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

Setting these two equal gives

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

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

${\displaystyle \tan {\frac {\phi -\psi }{2}}={\frac {k-1}{k+1}}\tan {\frac {\phi +\psi }{2}}.}$

This is the second equation we need. Once we solve the two equations for the two unknowns ${\displaystyle \phi }$ and ${\displaystyle \psi }$, we can use either of the two expressions above for ${\displaystyle {\frac {AB}{P_{1}P_{2}}}}$ to find P₁P₂ since AB is known. We can then find all the other segments using the law of sines.[1]

Solution algorithm

We are given four angles (α₁, β₁, α₂, β₂) and the distance AB. The calculation proceeds as follows:

• Calculate ${\displaystyle \gamma =\pi -\alpha _{1}-\beta _{1}-\beta _{2},\quad \delta =\pi -\alpha _{2}-\beta _{1}-\beta _{2}.}$
• Calculate ${\displaystyle k={\frac {\sin \gamma \sin \alpha _{2}\sin \beta _{1}}{\sin \delta \sin \alpha _{1}\sin \beta _{2}}}.}$
• Let ${\displaystyle s=\beta _{1}+\beta _{2},\quad d=2\arctan \left[{\frac {k-1}{k+1}}\tan(s/2)\right]}$ and then ${\displaystyle \phi =(s+d)/2,\quad \psi =(s-d)/2.}$
• Calculate

${\displaystyle P_{1}P_{2}=AB{\frac {\sin \phi \sin \delta }{\sin \alpha _{2}\sin \beta _{1}}}}$

or equivalently

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

If one of these fractions has a denominator close to zero, use the other one.

See also

• Solving triangles
• Snell's problem

References

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