Menu
Home
People
Places
Arts
History
Plants & Animals
Science
Life & Culture
Technology
Reference.org
Hansen's problem
open-in-new
<h2 id="solution-method-overview">Solution method overview</h2> <p>Define the following angles: γ = ∠ P 1 A P 2 , δ = ∠ P 1 B P 2 , ϕ = ∠ P 2 A B , ψ = ∠ P 1 B A . {\displaystyle {\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 <i>β</i>1 and <i>β</i>2, yielding the equation </p><p> ϕ + ψ = β 1 + β 2 . {\displaystyle \phi +\psi =\beta _{1}+\beta _{2}.} </p><p>A second equation can be found more laboriously, as follows. The <a href="/facts/Law_of_sines/PXNDVjlN">law of sines</a> yields </p><p> A B ¯ P 2 B ¯ = sin α 2 sin ϕ , P 2 B ¯ P 1 P 2 ¯ = sin β 1 sin δ . {\displaystyle {\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 }}.} </p><p>Combining these, we get </p><p> A B ¯ P 1 P 2 ¯ = sin α 2 sin β 1 sin ϕ sin δ . {\displaystyle {\frac {\overline {AB}}{\overline {P_{1}P_{2}}}}={\frac {\sin \alpha _{2}\sin \beta _{1}}{\sin \phi \sin \delta }}.} </p><p>Entirely analogous reasoning on the other side yields </p><p> A B ¯ P 1 P 2 ¯ = sin α 1 sin β 2 sin ψ sin γ . {\displaystyle {\frac {\overline {AB}}{\overline {P_{1}P_{2}}}}={\frac {\sin \alpha _{1}\sin \beta _{2}}{\sin \psi \sin \gamma }}.} </p><p>Setting these two equal gives </p><p> sin ϕ sin ψ = sin γ sin α 2 sin β 1 sin δ sin α 1 sin β 2 = k . {\displaystyle {\frac {\sin \phi }{\sin \psi }}={\frac {\sin \gamma \sin \alpha _{2}\sin \beta _{1}}{\sin \delta \sin \alpha _{1}\sin \beta _{2}}}=k.} </p><p>Using a known <a href="/facts/Trigonometric_identity/Xw9Jx8Vz">trigonometric identity</a> this ratio of sines can be expressed as the tangent of an angle difference: </p> tan 1 2 ( ϕ − ψ ) = k − 1 k + 1 tan 1 2 ( ϕ + ψ ) . {\displaystyle \tan {\tfrac {1}{2}}(\phi -\psi )={\frac {k-1}{k+1}}\tan {\tfrac {1}{2}}(\phi +\psi ).} <p>Where k = sin ϕ sin ψ . {\displaystyle k={\frac {\sin \phi }{\sin \psi }}.} </p><p>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 A B ¯ P 1 P 2 ¯ {\displaystyle {\tfrac {\overline {AB}}{\overline {P_{1}P_{2}}}}} to find P 1 P 2 ¯ {\displaystyle {\overline {P_{1}P_{2}}}} since AB is known. We can then find all the other segments using the law of sines.<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a> </p> <h2 id="solution-algorithm">Solution algorithm</h2> <p>We are given four angles (<i>α</i>1, <i>β</i>1, <i>α</i>2, <i>β</i>2) and the distance AB. The calculation proceeds as follows: </p> <ul><li>Calculate γ = π − α 1 − β 1 − β 2 , δ = π − α 2 − β 1 − β 2 . {\displaystyle {\begin{aligned}\gamma &=\pi -\alpha _{1}-\beta _{1}-\beta _{2},\\\delta &=\pi -\alpha _{2}-\beta _{1}-\beta _{2}.\end{aligned}}} </li> <li>Calculate k = sin γ sin α 2 sin β 1 sin δ sin α 1 sin β 2 . {\displaystyle k={\frac {\sin \gamma \sin \alpha _{2}\sin \beta _{1}}{\sin \delta \sin \alpha _{1}\sin \beta _{2}}}.} </li> <li>Let s = β 1 + β 2 , d = 2 arctan ( k − 1 k + 1 tan 1 2 s ) {\displaystyle s=\beta _{1}+\beta _{2},\quad d=2\arctan \left({\frac {k-1}{k+1}}\tan {\tfrac {1}{2}}s\right)} and then ϕ = s + d 2 , ψ = s − d 2 . {\displaystyle \phi ={\frac {s+d}{2}},\quad \psi ={\frac {s-d}{2}}.} </li></ul> <p>Calculate P 1 P 2 ¯ = A B ¯ sin ϕ sin δ sin α 2 sin β 1 {\displaystyle {\overline {P_{1}P_{2}}}={\overline {AB}}\ {\frac {\sin \phi \sin \delta }{\sin \alpha _{2}\sin \beta _{1}}}} or equivalently P 1 P 2 ¯ = A B ¯ sin ψ sin γ sin α 1 sin β 2 . {\displaystyle {\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. </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Solving_triangles/Dx8AK9bp">Solving triangles</a></li> <li><a href="/facts/Snellius%25E2%2580%2593Pothenot_problem/UMHQ2wRk">Snell's problem</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Udo Hebisch: Ebene und Sphaerische Trigonometrie, Kapitel 1, Beispiel 4 (2005, 2006)[1] Archived 2016-02-22 at the Wayback Machine <a href="http://www.mathe.tu-freiberg.de/~hebisch/skripte/sphaerik/strigo.pdf" target="_blank">http://www.mathe.tu-freiberg.de/~hebisch/skripte/sphaerik/strigo.pdf</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> </ol>