Snellius–Pothenot problem

Given are two lengths AC and BC, and three angles ${\displaystyle \alpha }$, ${\displaystyle \beta }$ and C, the solution proceeds as follows.

• calculate ${\displaystyle \phi =\operatorname {atan2} ({\rm {BC}}\sin \alpha ,{\rm {AC}}\sin \beta )}$. Where atan2 is a computer function, also called the arctangent of two arguments, that returns the arctangent of the ratio of the two values given. Note that in Microsoft Excel the two arguments are reversed, so the proper syntax would be '${\displaystyle ={\rm {atan2(AC^{*}\backslash \sin(beta),BC^{*}\backslash \sin(alpha))}}}$'. The atan2 function correctly handles the case where one of the two arguments is zero.
• calculate ${\displaystyle K=2\pi -\alpha -\beta -C.}$
• calculate ${\displaystyle W=2\cdot \operatorname {atan} \left[\tan(\pi /4-\phi )\tan \left({\frac {1}{2}}(\alpha +\beta +C)\right)\right].}$
• find ${\displaystyle x=(K+W)/2}$ and ${\displaystyle y=(K-W)/2.}$
• if ${\displaystyle |\sin \beta |>|\sin \alpha |}$ calculate ${\displaystyle {\rm {PC}}={\frac {{\rm {BC}}\sin y}{\sin \beta }}}$ else use ${\displaystyle {\rm {PC}}={\frac {{\rm {AC}}\sin x}{\sin \alpha }}.}$
• find ${\displaystyle {\rm {PA}}={\sqrt {{\rm {AC}}^{2}+{\rm {PC}}^{2}-2\cdot {\rm {AC}}\cdot {\rm {PC}}\cdot \cos(\pi -\alpha -x)}}.}$ (This comes from the law of cosines.)
• find ${\displaystyle {\rm {PB}}={\sqrt {{\rm {BC}}^{2}+{\rm {PC}}^{2}-2\cdot {\rm {BC}}\cdot {\rm {PC}}\cdot \cos(\pi -\beta -y)}}.}$

If the coordinates of ${\displaystyle A:x_{A},y_{A}}$ and ${\displaystyle C:x_{C},y_{C}}$ are known in some appropriate Cartesian coordinate system then the coordinates of ${\displaystyle P}$ can be found as well.

By the inscribed angle theorem the locus of points from which AC subtends an angle ${\displaystyle \alpha }$ is a circle having its center on the midline of AC; from the center O of this circle AC subtends an angle ${\displaystyle 2\alpha }$. Similarly the locus of points from which CB subtends an angle ${\displaystyle \beta }$ is another circle. The desired point P is at the intersection of these two loci.

Therefore, on a map or nautical chart showing the points A, B, C, the following graphical construction can be used:

• Draw the segment AC, the midpoint M and the midline, which crosses AC perpendicularly at M. On this line find the point O such that ${\displaystyle MO={\frac {AC}{2\tan \alpha }}}$. Draw the circle with center at O passing through A and C.
• Repeat the same construction with points B, C and the angle ${\displaystyle \beta }$.
• Mark P at the intersection of the two circles (the two circles intersect at two points; one intersection point is C and the other is the desired point P.)

This method of solution is sometimes called Cassini's method.

The following solution is based upon a paper by N. J. Wildberger.[2] It has the advantage that it is almost purely algebraic. The only place trigonometry is used is in converting the angles to spreads. There is only one square root required.

• define the following:
  • ${\displaystyle s(x)=\sin ^{2}(x)}$
  • ${\displaystyle A(x,y,z)=(x+y+z)^{2}-2(x^{2}+y^{2}+z^{2})}$
  • ${\displaystyle r_{1}=s(\beta )}$
  • ${\displaystyle r_{2}=s(\alpha )}$
  • ${\displaystyle r_{3}=s(\alpha +\beta )}$
  • ${\displaystyle Q_{1}=BC^{2}}$
  • ${\displaystyle Q_{2}=AC^{2}}$
  • ${\displaystyle Q_{3}=AB^{2}}$
• now let:
  • ${\displaystyle R_{1}=r_{2}Q_{3}/r_{3}}$
  • ${\displaystyle R_{2}=r_{1}Q_{3}/r_{3}}$
  • ${\displaystyle C_{0}=((Q_{1}+Q_{2}+Q_{3})(r_{1}+r_{2}+r_{3})-2(Q_{1}r_{1}+Q_{2}r_{2}+Q_{3}r_{3}))/(2r_{3})}$
  • ${\displaystyle D_{0}=r_{1}r_{2}A(Q_{1},Q_{2},Q_{3})/r_{3}}$
• the following equation gives two possible values for ${\displaystyle R_{3}}$:
  • ${\displaystyle (R_{3}-C_{0})^{2}=D_{0}}$
• choosing the larger of these values, let:
  • ${\displaystyle v_{1}=1-(R_{1}+R_{3}-Q_{2})^{2}/(4R_{1}R_{3})}$
  • ${\displaystyle v_{2}=1-(R_{2}+R_{3}-Q_{1})^{2}/(4R_{2}R_{3})}$
• finally:
  • ${\displaystyle AP^{2}=v_{1}R_{1}/r_{2}=v_{1}Q_{3}/r_{3}}$
  • ${\displaystyle BP^{2}=v2R_{2}/r_{1}=v_{2}Q_{3}/r_{3}}$

When the point P happens to be located on the same circle as A, B and C, the problem has an infinite number of solutions; the reason is that from any other point P' located on the arc APB of this circle the observer sees the same angles alpha and beta as from P (inscribed angle theorem). Thus the solution in this case is not uniquely determined.

The circle through ABC is known as the "danger circle", and observations made on (or very close to) this circle should be avoided. It is helpful to plot this circle on a map before making the observations.

A theorem on cyclic quadrilaterals is helpful in detecting the indeterminate situation. The quadrilateral APBC is cyclic iff a pair of opposite angles (such as the angle at P and the angle at C) are supplementary i.e. iff ${\displaystyle \alpha +\beta +C=k\pi ,(k=1,2,\cdots )}$. If this condition is observed the computer/spreadsheet calculations should be stopped and an error message ("indeterminate case") returned.

(Adapted form Bowser,[3] exercise 140, page 203). A, B and C are three objects such that AC = 435 (yards), CB = 320, and C = 255.8 degrees. From a station P it is observed that APC = 30 degrees and CPB = 15 degrees. Find the distances of P from A, B and C. (Note that in this case the points C and P are on the same side of the line AB, a different configuration from the one shown in the figure).

Answer: PA = 790, PB = 777, PC = 502.

A slightly more challenging test case for a computer program uses the same data but this time with CPB = 0. The program should return the answers 843, 1157 and 837.

Plaque on Snellius' house in Leiden

The British authority on geodesy, George Tyrrell McCaw (1870–1942) wrote that the proper term in English was Snellius problem, while Snellius-Pothenot was the continental European usage.[4]

McCaw thought the name of Laurent Pothenot (1650–1732) did not deserve to be included as he had made no original contribution, but merely restated Snellius 75 years later.

• Solution of triangles
• Triangulation (surveying)

• Gerhard Heindl: Analysing Willerding’s formula for solving the planar three point resection problem, Journal of Applied Geodesy, Band 13, Heft 1, Seiten 27–31, ISSN (Online) 1862-9024, ISSN (Print) 1862-9016, DOI:

• Edward A. Bowser: A treatise on plane and spherical trigonometry, Washington D.C., Heath & Co., 1892, page 188 Google books