View source for Solution of triangles

{{short description|Problem of finding unknown lengths and angles of a triangle}}

'''Solution of triangles''' ({{lang-la|solutio triangulorum}}) is the main [[trigonometry|trigonometric]] problem of finding the characteristics of a [[triangle]] (angles and lengths of sides), when some of these are known. The triangle can be located on a [[plane (geometry)|plane]] or on a [[sphere]]. Applications requiring triangle solutions include [[geodesy]], [[astronomy]], [[construction]], and [[navigation]].

==Solving plane triangles==
[[File:Triangle - angles, vertices, sides.svg|thumb|220px|right|{{center|Standard notation for a triangle}}]]
A general form triangle has six main characteristics (see picture): three linear (side lengths {{mvar|a, b, c}}) and three angular ({{mvar|α, β, γ}}). The classical plane trigonometry problem is to specify three of the six characteristics and  determine the other three.  A triangle can be uniquely determined in this sense when given any of the following:<ref>{{cite web |url=http://www.mathsisfun.com/algebra/trig-solving-triangles.html |title=Solving Triangles |publisher=Maths is Fun |access-date=4 April 2012<!-- 6:37 (UTC)-->}}</ref><ref>{{cite web |url=http://web.horacemann.org/academics/math/pcbch/trig/triangle.html |title=Solving Triangles |publisher=web.horacemann.org |access-date=4 April 2012<!-- 6:42 (UTC)--> |url-status=dead|archive-url=https://web.archive.org/web/20140107224746/http://web.horacemann.org/academics/math/pcbch/trig/triangle.html |archive-date=7 January 2014 }}</ref>

*Three sides ('''SSS''')
*Two sides and the included angle ('''SAS''', side-angle-side)
*Two sides and an angle not included between them ('''SSA'''), if the side length adjacent to the angle is shorter than the other side length.
*A side and the two angles adjacent to it ('''ASA''')
*A side, the angle opposite to it and an angle adjacent to it ('''AAS''').

For all cases in the plane, at least one of the side lengths must be specified. If only the angles are given, the side lengths cannot be determined, because any [[similarity (geometry)|similar]] triangle is a solution.

===Trigonomic relations===
[[File:Beliebiges Dreieck cen.png|thumb|upright=2.0|Overview of particular steps and tools used when solving plane triangles]]
The standard method of solving the problem is to use fundamental relations.
;[[Law of cosines]]<math display=block>\begin{align}
  a^2 &= b^2 + c^2 - 2 b c \cos \alpha \\
  b^2 &= a^2 + c^2 - 2 a c \cos \beta \\
  c^2 &= a^2 + b^2 - 2 a b \cos \gamma 
\end{align}</math>

;[[Law of sines]]<math display=block>
  \frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}</math>
;[[Sum of angles of a triangle|Sum of angles]]:<math display=block>
  \alpha + \beta + \gamma = 180^\circ 
</math>
;[[Law of tangents]]<math display=block>\frac{a-b}{a+b} = \frac{\tan\frac12(\alpha-\beta)}{\tan\tfrac12 (\alpha+\beta)}.</math>
There are other (sometimes practically useful) universal relations: the [[law of cotangents]] and [[Mollweide's formula]].

====Notes====
#To find an unknown angle, the [[law of cosines]] is safer than the [[law of sines]]. The reason is that the value of [[sine]] for the angle of the triangle does not uniquely determine this angle. For example, if {{math|sin ''β'' {{=}} 0.5}}, the angle {{mvar|β}} can equal either 30° or 150°. Using the law of cosines avoids this problem: within the interval from 0° to 180° the cosine value unambiguously determines its angle. On the other hand, if the angle is small (or close to 180°), then it is more robust numerically to determine it from its sine than its cosine because the arc-cosine function has a divergent derivative at 1 (or −1).
#We assume that the relative position of specified characteristics is known. If not, the mirror reflection of the triangle will also be a solution. For example, three side lengths uniquely define either a triangle or its reflection.

===Three sides given (SSS)===
[[File:resolve triangle with a b c.png|thumb|right|250px|{{center|Three sides given}}]]

Let three side lengths {{mvar|a, b, c}} be specified. To find the angles {{mvar|α, β}}, the [[law of cosines]] can be used:<ref>{{cite web |url=http://www.mathsisfun.com/algebra/trig-solving-sss-triangles.html |title=Solving SSS Triangles |publisher=Maths is Fun |access-date=13 January 2015}}</ref>
<math display=block>\begin{align}
  \alpha &= \arccos \frac{b^2 + c^2 - a^2}{2bc} \\[4pt]
  \beta &= \arccos \frac{a^2 + c^2 - b^2}{2ac}.
\end{align}</math>

Then angle {{math|''γ'' {{=}} 180° − ''α'' − ''β''}}.

Some sources recommend to find angle {{mvar|β}} from the [[law of sines]] but (as Note 1 above states) there is a risk of confusing an acute angle value with an obtuse one.

Another method of calculating the angles from known sides is to apply the [[law of cotangents]].

Area using [[Heron's formula]]:
<math> A = \sqrt{s(s-a)(s-b)(s-c)}</math> where <math> s= \frac{a+b+c}{2} </math>

Área using Jesús Sánchez formula:
<math>
  A = \frac{\sqrt{(a+b+c)(b+c-a)(a^2-(b-c)^2)}}{4}
</math>

===Two sides and the included angle given (SAS)===
[[File:resolve triangle with a b gamma.png|thumb|right|250px|{{center|Two sides and the included angle given}}]]

Here the lengths of sides {{mvar|a, b}} and the angle {{mvar|γ}} between these sides are known. The third side can be determined from the law of cosines:<ref>{{cite web |url=http://www.mathsisfun.com/algebra/trig-solving-sas-triangles.html |title=Solving SAS Triangles |publisher=Maths is Fun |access-date=13 January 2015}}</ref>
<math display=block>
  c = \sqrt{a^2+b^2-2ab\cos\gamma}.
</math>
Now we use law of cosines to find the second angle:
<math display=block> 
  \alpha =  \arccos \frac{b^2 + c^2 - a^2}{2bc}.
</math>
Finally, {{math|''β'' {{=}} 180° − ''α'' − ''γ''}}.

===Two sides and non-included angle given (SSA)===
[[File:resolve triangle with b c beta.png|thumb|right|250px|{{center|Two sides and a non-included angle given}}]]
[[File:Resolve triangle with b c beta 2 solutions.png|thumb|250px|{{center|Two solutions for the triangle}}]]

This case is not solvable in all cases; a solution is guaranteed to be unique only if the side length adjacent to the angle is shorter than the other side length.  Assume that two sides {{mvar|b, c}} and the angle {{mvar|β}} are known. The equation for the angle {{mvar|γ}} can be implied  from the [[law of sines]]:<ref>{{cite web |url=http://www.mathsisfun.com/algebra/trig-solving-ssa-triangles.html |title=Solving SSA Triangles |publisher=Maths is Fun |access-date=9 March 2013}}</ref>
<math display=block>\sin\gamma = \frac c b \sin\beta.</math>
We denote further {{math|''D'' {{=}} {{sfrac|''c''|''b''}} sin ''β''}} (the equation's right side). There are four possible cases:
#If {{math|''D'' > 1}}, no such triangle exists because the side {{mvar|b}} does not reach line {{mvar|BC}}. For the same reason a solution does not exist if the angle {{math|''β'' ≥ 90°}} and {{math|''b'' ≤ ''c''}}.
#If {{math|''D'' {{=}} 1}}, a unique solution exists: {{math|''γ'' {{=}} 90°}}, i.e., the triangle is [[right triangle|right-angled]].
# If {{math|''D'' < 1}} two alternatives are possible.
## If {{math|''b'' ≥ ''c''}}, then {{math|''β'' ≥ ''γ''}} (the larger side corresponds to a larger angle). Since no triangle can have two obtuse angles, {{math|''γ''}} is an acute angle and the solution {{math|''γ'' {{=}} arcsin ''D''}} is unique.
## If {{math|''b'' < ''c''}}, the angle {{math|''γ''}} may be acute: {{math|''γ'' {{=}} arcsin ''D''}} or obtuse: {{math|''{{prime|γ}}'' {{=}} 180° − ''γ''}}. The figure on right shows the point {{math|''C''}}, the side {{math|''b''}} and the angle {{math|''γ''}} as the first solution, and the point {{math|''{{prime|C}}''}}, side {{math|''{{prime|b}}''}} and the angle {{math|''{{prime|γ}}''}} as the second solution.

Once {{math|''γ''}} is obtained, the third angle {{math|''α'' {{=}} 180° − ''β'' − ''γ''}}.

The third side can then be found from the law of sines:
<math display=block>a = b\ \frac{\sin\alpha}{\sin\beta}</math>

or from the law of cosines:
<math display=block>a = c\cos\beta \pm  \sqrt{b^2 -c^2\sin^2\beta} </math>

===A side and two adjacent angles given (ASA)===
[[File:resolve triangle with c alpha beta.png|thumb|right|250px|{{center|One side and two adjacent angles given}}]]

The known characteristics are the side  {{mvar|c}} and the angles {{mvar|α, β}}. The third angle {{math|''γ'' {{=}} 180° − ''α'' − ''β''}}.

Two unknown sides can be calculated from the law of sines:<ref>{{cite web |url=http://www.mathsisfun.com/algebra/trig-solving-asa-triangles.html |title=Solving ASA Triangles |publisher=Maths is Fun |access-date=13 January 2015}}</ref>
<math display=block>\begin{align}
  a &= c\ \frac{\sin\alpha}{\sin\gamma} = c\ \frac{\sin\alpha}{\sin\alpha \cos \beta + \sin\beta \cos \alpha} \\[4pt]
  b &= c\ \frac{\sin\beta}{\sin\gamma} = c\ \frac{\sin\beta}{\sin\alpha \cos \beta + \sin\beta \cos \alpha}
\end{align}</math>

===A side, one adjacent angle and the opposite angle given (AAS)===
The procedure for solving an AAS triangle is same as that for an ASA triangle: First, find the third angle by using the angle sum property of a triangle, then find the other two sides using the [[law of sines]].

===Other given lengths===

In many cases, triangles can be solved given three pieces of information some of which are the lengths of the triangle's [[median (geometry)|medians]], [[altitude (geometry)|altitudes]], or [[angle bisector]]s. Posamentier and Lehmann<ref>Alfred S. Posamentier and Ingmar Lehmann, ''[[The Secrets of Triangles]]'', Prometheus Books, 2012: pp. 201–203.</ref> list the results for the question of solvability using no higher than square roots (i.e., [[Straightedge and compass construction|constructibility]]) for each of the 95 distinct cases; 63 of these are constructible.

==Solving spherical triangles==
[[File:Spherical triangle 3d.png|thumb|250 px|right|{{center|Spherical triangle}}]]

The general [[spherical triangle]] is fully determined by three of its six characteristics (3 sides and 3 angles). The lengths of the sides {{mvar|a, b, c}} of a spherical triangle are their [[central angle]]s, measured in angular units rather than linear units. (On a unit sphere, the angle (in [[radians]]) and length around the sphere are numerically the same. On other spheres, the angle (in radians) is equal to the length around the sphere divided by the radius.)

[[Spherical geometry]] differs from planar [[Euclidean geometry]], so the solution of spherical triangles is built on different rules. For example, the [[sum of angles of a triangle|sum of the three angles]] {{math|''α'' + ''β'' + ''γ''}} depends on the size of the triangle. In addition, [[similarity (geometry)|similar triangles]] cannot be unequal, so the problem of constructing a triangle with specified three angles has a unique solution. The basic relations used to solve a problem are similar to those of the planar case: see [[Spherical law of cosines]] and [[Spherical law of sines]].

Among other relationships that may be useful are the [[half-side formula]] and [[Napier's analogies]]:<ref>[http://mathworld.wolfram.com/NapiersAnalogies.html Napier's Analogies] at MathWorld</ref>
<math display=block>\begin{align}
  \tan\tfrac12 c \, \cos\tfrac12(\alpha-\beta) &= \tan\tfrac12(a+\,b) \cos\tfrac12(\alpha+\beta) \\
  \tan\tfrac12 c \, \sin\tfrac12(\alpha-\beta) &= \tan\tfrac12(a\ \!-\,b)  \sin\tfrac12(\alpha+\beta) \\
  \cot\tfrac12\gamma \ \! \cos\tfrac12(a\ \!-\,b) &= \tan\tfrac12(\alpha+\beta) \cos\tfrac12(a+b) \\
  \cot\tfrac12\gamma \, \sin\tfrac12(a\ \!-\,b) &= \tan\tfrac12(\alpha-\beta) \sin\tfrac12(a+b).
\end{align}</math>

[[File:Solve spherical triangle with a b c.png|thumb|right|250px|{{center|Three sides given}}]]

===Three sides given (spherical SSS)===
Known: the sides {{math|''a'', ''b'', ''c''}} (in angular units). The triangle's angles are computed using the [[spherical law of cosines]]:
<math display=block>\begin{align}
  \alpha &= \arccos\frac{\cos a-\cos b\ \cos c}{\sin b\ \sin c}, \\[4pt]
  \beta  &= \arccos\frac{\cos b-\cos c\ \cos a}{\sin c\ \sin a}, \\[4pt]
  \gamma &= \arccos\frac{\cos c-\cos a\ \cos b}{\sin a\ \sin b}.
\end{align}</math>

{{Clear}}
[[File:Solve spherical triangle with a b gamma.png|thumb|right|250px|{{center|Two sides and the included angle given}}]]

===Two sides and the included angle given (spherical SAS)===
Known: the sides {{mvar|a, b}} and the angle {{mvar|γ}} between them. The side {{mvar|c}} can be found from the [[spherical law of cosines]]:
<math display=block>c = \arccos \left(\cos a\cos b + \sin a\sin b\cos\gamma \right).</math>

The angles {{mvar|α, β}} can be calculated as above, or by using Napier's analogies:

<math display=block>\begin{align}
  \alpha &= \arctan\ \frac{2\sin a}{\tan\frac12\gamma \, \sin(b+a) + \cot\frac12\gamma \, \sin(b-a)}, \\[4pt]
  \beta  &= \arctan\ \frac{2\sin b}{\tan\frac12\gamma \, \sin(a+b) + \cot\frac12\gamma \, \sin(a-b)}.
\end{align}</math>

This problem arises in the [[great-circle navigation#Course|navigation problem]] of finding the great circle between two points on the earth specified by their latitude and longitude; in this application, it is important to use formulas which are not susceptible to round-off errors.  For this purpose, the following formulas (which may be derived using vector algebra) can be used:
<math display=block>\begin{align}
  c &= \arctan\frac{\sqrt{(\sin a\cos b - \cos a \sin b \cos \gamma)^2 + (\sin b\sin\gamma)^2}}{\cos a \cos b + \sin a\sin b\cos\gamma}, \\[4pt]
  \alpha &= \arctan\frac{\sin a\sin\gamma}{\sin b\cos a - \cos b\sin a\cos\gamma}, \\[4pt]
  \beta &= \arctan\frac{\sin b\sin\gamma}{\sin a\cos b - \cos a\sin b\cos\gamma},
\end{align}</math>
where the signs of the numerators and denominators in these expressions should be used to determine the quadrant of the arctangent.

{{Clear}}
[[File:Solve spherical triangle with b c beta.png|thumb|right|250px|{{center|Two sides and a non-included angle given}}]]

===Two sides and non-included angle given (spherical SSA)===
This problem is not solvable in all cases; a solution is guaranteed to be unique only if the side length adjacent to the angle is shorter than the other side length.  Known: the sides {{mvar|b, c}} and the angle {{mvar|β}} not between them. A solution exists if the following condition holds:
<math display=block>b > \arcsin\! \bigl(\sin c\,\sin\beta \bigr).</math>
The angle {{mvar|γ}} can be found from the [[spherical law of sines]]:
<math display=block>\gamma = \arcsin \frac{\sin c\,\sin\beta}{\sin b}.</math>
As for the plane case, if {{math|''b'' < ''c''}} then there are two solutions: {{mvar|γ}} and {{math|180° - ''γ''}}.

We can find other characteristics by using Napier's analogies:
<math display=block>\begin{align}
  a &= 2\arctan \left[ \tan\tfrac12(b-c) \ \frac{\sin\tfrac12(\beta+\gamma)}{\sin\tfrac12(\beta-\gamma)} \right], \\[4pt]
  \alpha &= 2\arccot \left[\tan\tfrac12(\beta-\gamma) \ \frac{\sin \tfrac12(b+c)}{\sin \tfrac12(b-c)} \right].
\end{align}</math>

[[File:Solve spherical triangle with c alpha beta.png|thumb|right|250px|{{center|One side and two adjacent angles given}}]]

===A side and two adjacent angles given (spherical ASA)===
Known: the side {{mvar|c}} and the angles {{mvar|α, β}}. First we determine the angle {{mvar|γ}} using the [[spherical law of cosines]]:
<math display=block>\gamma = \arccos\! \bigl(\sin\alpha\sin\beta\cos c -\cos\alpha\cos\beta \bigr).\,</math>

We can find the two unknown sides from the spherical law of cosines (using the calculated angle {{mvar|γ}}):

<math display=block>\begin{align}
  a &= \arccos\frac{\cos\alpha+\cos\beta\cos\gamma}{\sin\beta\sin\gamma}, \\[4pt]
  b &= \arccos \frac{\cos\beta+\cos\alpha\cos\gamma}{\sin\alpha\sin\gamma},
\end{align}</math>

or by using Napier's analogies:
<math display=block>\begin{align}
  a &= \arctan\frac{2\sin\alpha}{\cot\frac12 c \, \sin(\beta+\alpha) + \tan\frac12 c \, \sin(\beta-\alpha)}, \\[4pt]
  b &= \arctan\frac{2\sin\beta} {\cot\frac12 c \, \sin(\alpha+\beta) + \tan\frac12 c \, \sin(\alpha-\beta)}.
\end{align}</math>

{{Clear}}
[[File:Solve spherical triangle with a alpha beta.png|thumb|right|250px|{{center|One side, one adjacent angle and the opposite angle given}}]]

===A side, one adjacent angle and the opposite angle given (spherical AAS)===
Known: the side {{mvar|a}} and the angles {{mvar|α, β}}. The side {{mvar|b}} can be found from the [[spherical law of sines]]:
<math display=block>b = \arcsin  \frac{\sin a\,\sin \beta}{\sin \alpha} .</math>

If the angle for the side {{math|''a''}} is acute and {{math|''α'' > ''β''}}, another solution exists:
<math display=block>b = \pi - \arcsin \frac{\sin a\,\sin \beta}{\sin \alpha} .</math>

We can find other characteristics by using Napier's analogies:
<math display=block>\begin{align}
  c &= 2\arctan \left[ \tan\tfrac12(a-b) \ \frac{\sin\tfrac12(\alpha+\beta)}{\sin\frac12(\alpha-\beta)}\right], \\[4pt]
  \gamma &= 2\arccot \left[\tan\tfrac12(\alpha-\beta) \ \frac{\sin \tfrac12(a+b)}{\sin \frac12(a-b)} \right].
\end{align}</math>

{{Clear}}
[[File:Solve spherical triangle with alpha beta gamma.png|thumb|right|250px|{{center|Three angles given}}]]

===Three angles given (spherical AAA)===
Known: the angles {{mvar|α, β, γ}}. From the [[spherical law of cosines]] we infer:
<math display=block>\begin{align}
  a &= \arccos\frac{\cos\alpha + \cos\beta\cos\gamma}{\sin\beta\sin\gamma}, \\[4pt]
  b &= \arccos\frac{\cos\beta + \cos\gamma\cos\alpha}{\sin\gamma\sin\alpha}, \\[4pt]
  c &= \arccos\frac{\cos\gamma + \cos\alpha\cos\beta}{\sin\alpha\sin\beta}.
\end{align}</math>

===Solving right-angled spherical triangles===
The above algorithms become much simpler if one of the angles of a triangle (for example, the angle {{mvar|C}}) is the right angle. Such a spherical triangle is fully defined by its two elements, and the other three can be calculated using [[spherical trigonometry#Napier's Pentagon|Napier's Pentagon]] or the following relations.
:<math>\sin a = \sin c \cdot \sin A</math> (from the [[spherical law of sines]])
:<math>\tan a = \sin b \cdot \tan A</math>
:<math>\cos c = \cos a \cdot \cos b</math> (from the [[spherical law of cosines]])
:<math>\tan b = \tan c \cdot \cos A</math>
:<math>\cos A = \cos a \cdot \sin B</math> (also from the spherical law of cosines)
:<math>\cos c = \cot A \cdot \cot B</math>

==Some applications==

===Triangulation===
[[File:distance by triangulation.svg|thumb|right|320px|{{center|Distance measurement by [[triangulation]]}}]]
{{Main|Triangulation}}
If one wants to measure the distance {{mvar|d}} from shore to a remote ship via triangulation, one marks on the shore two points with known distance {{mvar|l}} between them (the baseline). Let {{mvar|α, β}} be the angles between the baseline and the direction to the ship.

From the formulae above (ASA case, assuming planar geometry) one can compute the distance as the [[altitude (triangle)|triangle height]]:
<math display=block>
  d = \frac{\sin\alpha\,\sin\beta}{\sin(\alpha + \beta)} \ell = \frac{\tan\alpha\,\tan\beta}{\tan\alpha + \tan\beta} \ell.</math>

For the spherical case, one can first compute the length of side from the point at {{mvar|α}} to the ship (i.e. the side opposite to {{mvar|β}}) via the ASA formula
<math display=block> 
  \tan b = \frac{2\sin\beta}{\cot\frac12 \ell \, \sin(\alpha + \beta) + \tan\frac12 \ell \, \sin(\alpha - \beta)},
</math>
and insert this into the AAS formula for the right subtriangle that contains the angle {{mvar|α}} and the sides {{mvar|b}} and {{mvar|d}}:
<math display=block> 
  \sin d = \sin b \sin\alpha = \frac{\tan b}{\sqrt{1 + \tan^2 b}}\sin\alpha.
</math>
(The planar formula is actually the first term of the Taylor expansion of {{mvar|d}} of the spherical solution in powers of {{mvar|ℓ}}.)

This method is used in [[cabotage]]. The angles {{mvar|α, β}} are defined  by observation of familiar landmarks from the ship.
{{Clear}}
[[File:mountain height by triangulation.svg|thumb|right|320px|{{center|How to measure a mountain's height}}]]
As another example, if one wants to measure the height {{mvar|h}} of a mountain or a high building, the angles {{mvar|α, β}} from two ground points to the top are specified. Let {{math|''ℓ''}} be the distance between these points. From the same ASA case formulas we obtain:
<math display=block> 
  h = \frac{\sin\alpha\,\sin\beta}{\sin(\beta-\alpha)} \ell = \frac{\tan\alpha\,\tan\beta}{\tan\beta-\tan\alpha} \ell.
</math>

===The distance between two points on the globe===
{{Main|Great-circle distance}}

[[File:distance on earth.png|224px|right]]
To calculate the distance between two points on the globe,
:Point A: latitude {{mvar|λ{{sub|A}}}}, longitude {{mvar|L{{sub|A}}}}, and
:Point B: latitude {{mvar|λ{{sub|B}}}}, longitude {{mvar|L{{sub|B}}}}
we consider the spherical triangle {{mvar|ABC}}, where {{mvar|C}} is the North Pole. Some characteristics are:
<math display=block>\begin{align}
  a &= 90^\circ - \lambda_B, \\
  b &= 90^\circ - \lambda_A, \\
  \gamma &= L_A - L_B.
\end{align}</math>
If [[#Two sides and the included angle given|two sides and the included angle given]], we obtain from the formulas
<math display=block>
 \overline{AB} = R \arccos\!\Bigr[\sin \lambda_A \sin \lambda_B + \cos \lambda_A \cos \lambda_B \cos(L_A - L_B)\Bigr].
</math>
Here {{mvar|R}} is the [[Earth|Earth's radius]].

==See also==
*[[Congruence (geometry)|Congruence]]
*[[Hansen's problem]]
*[[Hinge theorem]]
*[[Lénárt sphere]]
*[[Snellius–Pothenot problem]]

==References==
{{Reflist}}
*{{cite book |author=Euclid |author-link=Euclid |editor=Sir Thomas Heath |editor-link=Thomas Little Heath |title=The Thirteen Books of the Elements. Volume I |others=Translated with introduction and commentary |publisher=Dover |year=1956 |orig-year=1925 |isbn=0-486-60088-2 |url-access=registration |url=https://archive.org/details/thirteenbooksofe00eucl }}

==External links==
*[https://web.archive.org/web/20040404234808/http://www.pupress.princeton.edu/books/maor/ Trigonometric Delights], by [[Eli Maor]], Princeton University Press, 1998.  Ebook version, in PDF format, full text presented.
*[https://books.google.com/books/about/Trigonometry.html?id=RsxHAAAAIAAJ Trigonometry] by Alfred Monroe Kenyon and Louis Ingold, The Macmillan Company, 1914. In images, full text presented. Google book.
*''[http://mathworld.wolfram.com/SphericalTrigonometry.html Spherical trigonometry]'' on Math World.
*[http://www.rwgrayprojects.com/rbfnotes/trig/strig/strig.html Intro to Spherical Trig.] Includes discussion of The Napier circle and Napier's rules
*[https://archive.org/details/sphericaltrigono19770gut Spherical Trigonometry &mdash; for the use of colleges and schools] by I. Todhunter, M.A., F.R.S. Historical Math Monograph posted by [http://historical.library.cornell.edu/math/index.html Cornell University Library].
*''[http://www.rlefebvre.ca/triangulateur/triangulator.htm Triangulator]'' &ndash; Triangle solver. Solve any plane triangle problem with the minimum of input data. Drawing of the solved triangle.
*''[http://gnomonique.fr/trisph/index_en.htm TriSph]'' &ndash; Free software to solve the spherical triangles, configurable to different practical applications and configured for gnomonic.
*''[https://web.archive.org/web/20141009115504/https://files.nyu.edu/emf202/public/js/spherical.html  Spherical Triangle Calculator]'' &ndash; Solves spherical triangles.
*''[http://triancal.x10.bz TrianCal]'' &ndash; Triangles solver by [[Jesus S.]]


[[Category:Spherical trigonometry]]
[[Category:Triangle problems]]
[[Category:Trigonometry]]

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