Hansen's problem

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 Template:Mvar, and two unknown points Template:Math. From Template:Math and Template:Math 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 Template:Math and Template:Math. See figure; the angles measured are Template:Math.

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: $$\begin{array}{cccc}\gamma & =\mathrm{\angle }{P}_{1}A{P}_2,& \delta & =\mathrm{\angle }{P}_{1}B{P}_2,\\ \varphi & =\mathrm{\angle }{P}_{2}AB,& \psi & =\mathrm{\angle }{P}_{1}BA.\end{array}$$ As a first step we will solve for Template:Mvar and Template:Mvar. The sum of these two unknown angles is equal to the sum of Template:Math and Template:Math, yielding the equation

$$\varphi +\psi ={\beta }_1+{\beta }_2.$$

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

$$\frac{\overline{AB}}{\overline{P_{2}B}}=\frac{\sin {\alpha }_2}{\sin \varphi },\frac{\overline{P_{2}B}}{\overline{P_1{P}_2}}=\frac{\sin {\beta }_1}{\sin \delta }.$$

Combining these, we get

$$\frac{\overline{AB}}{\overline{P_1{P}_2}}=\frac{\sin {\alpha }_2\sin {\beta }_1}{\sin \varphi \sin \delta }.$$

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

$$\frac{\sin \varphi }{\sin \psi }=\frac{\sin \gamma \sin {\alpha }_2\sin {\beta }_1}{\sin \delta \sin {\alpha }_1\sin {\beta }_2}=k.$$

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

$\tan \frac{1}{2}(\varphi -\psi )=\frac{k-1}{k+1}\tan \frac{1}{2}(\varphi +\psi ).$

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

This is the second equation we need. Once we solve the two equations for the two unknowns Template:Mvar, we can use either of the two expressions above for $\frac{\overline{AB}}{\overline{P_1{P}_2}}$ to find Template:Tmath since Template:Overline is known. We can then find all the other segments using the law of sines.

Solution algorithm

We are given four angles Template:Math and the distance Template:Overline. The calculation proceeds as follows:

• Calculate $$\begin{array}{cc}\gamma & =\pi -{\alpha }_1-{\beta }_1-{\beta }_2,\\ \delta & =\pi -{\alpha }_2-{\beta }_1-{\beta }_2.\end{array}$$
• Calculate $$k=\frac{\sin \gamma \sin {\alpha }_2\sin {\beta }_1}{\sin \delta \sin {\alpha }_1\sin {\beta }_2}.$$
• Let $$s={\beta }_1+{\beta }_2,d=2\arctan (\frac{k-1}{k+1}\tan \frac{1}{2}s)$$ and then $$\varphi =\frac{s+d}{2},\psi =\frac{s-d}{2}.$$

Calculate $$\overline{P_1{P}_2}=\overline{AB}\frac{\sin \varphi \sin \delta }{\sin {\alpha }_2\sin {\beta }_1}$$ 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.

Solutions via Geometric Algebra

In addition to presenting algorithms for solving the problem via Vector Geometric Algebra and Conformal Geometric Algebra, Ventura et al. review previous methods, and compare the various methods' computational speeds and sensitivity to measurement error.

See also

• Solving triangles
• Snell's problem