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Solution method overview

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 β1 and β2, yielding the equation

ϕ + ψ = β 1 + β 2 . {\displaystyle \phi +\psi =\beta _{1}+\beta _{2}.}

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

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 }}.}

Combining these, we get

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 }}.}

Entirely analogous reasoning on the other side yields

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 }}.}

Setting these two equal gives

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.}

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

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 ).}

Where k = sin ⁡ ϕ sin ⁡ ψ . {\displaystyle 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 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.¹

Solution algorithm

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

• 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}}}
• 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}}}.}
• 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}}.}

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.

See also

• Solving triangles
• Snell's problem

References

1. Udo Hebisch: Ebene und Sphaerische Trigonometrie, Kapitel 1, Beispiel 4 (2005, 2006)[1] Archived 2016-02-22 at the Wayback Machine http://www.mathe.tu-freiberg.de/~hebisch/skripte/sphaerik/strigo.pdf ↩