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.

See also

• Solving triangles
• Snell's problem