Iterative refinement: Difference between revisions

Revision as of 16:00, 22 January 2014

Hello, I'm Casey, a 27 year old from Wroclaw, Poland.
My hobbies include (but are not limited to) Camping, Weightlifting and watching Modern Family.

Also visit my website ... Hostgator Discount Voucher Solution of triangles (Template:Lang-lat) is the historical term for solving the main 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 or on a sphere. Applications requiring triangle solutions include geodesy, astronomy, construction, and navigation.

Solving plane triangles

File:Triangle - angles, vertices, sides.svg

Standard notation in the triangle

A general form triangle has 6 main characteristics (see picture): 3 linear (side lengths $a, b, c$ ) and 3 angular ( $α, β, γ$ ). The classical plane trigonometry problem is to specify 3 of the 6 characteristics and determine the other three. At least one of the side lengths must be specified. If only the angles are given, the side lengths cannot be determined, because any similar triangle is a solution.

A triangle can be solved when given any of the following:^[1]^[2]

Three sides (SSS)
Two sides and the included angle (SAS)
Two sides and an angle not included between them (SSA)
A side and the two angles adjacent to it (ASA)
A side, the angle opposite to it and an angle adjacent to it (AAS).

Main theorems

File:Beliebiges Dreieck cen.png

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: $a^{2} = b^{2} + c^{2} - 2 b c \cos α$; $b^{2} = a^{2} + c^{2} - 2 a c \cos β$; $c^{2} = a^{2} + b^{2} - 2 a b \cos γ$
Law of sines: $\frac{a}{\sin α} = \frac{b}{\sin β} = \frac{c}{\sin γ}$
Sum of angles: $α + β + γ = 18 0^{\circ}$
Law of tangents: $\frac{a - b}{a + b} = \frac{t a n [\frac{1}{2} (α - β)]}{t a n [\frac{1}{2} (α + β)]} .$

There are other (sometimes practically useful) universal relations: the law of cotangents and Mollweide's formula.

Notes

To find an unknown angle, law of cosines is safer than 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 $\sin β = 0.5$ , the angle $β$ can be equal either $3 0^{\circ}$ or $15 0^{\circ}$ . Using the law of cosines avoids this problem: within the interval from $0^{\circ}$ to $18 0^{\circ}$ the cosine value determines its angle unambiguously. 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 be the solution also. For example, three side lengths uniquely define either a triangle or its reflection.

File:Resolve triangle with a b c.png

Three sides given

Three sides given (SSS)

Let three side lengths $a, b, c$ be specified. To find the angles $α, β$ , you can use the law of cosines:^[3]

α = \arccos \frac{b^{2} + c^{2} - a^{2}}{2 b c}

β = \arccos \frac{a^{2} + c^{2} - b^{2}}{2 a c} .

Then angle $γ = 18 0^{\circ} - α - β$ .

Some sources recommend to find angle $β$ from law of sines but (as Note 1 above states) there is a risk to confuse acute angle value with obtuse one.

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

File:Resolve triangle with a b gamma.png

Two sides and the included angle given

Two sides and the included angle given (SAS)

Here we know the lengths of sides $a, b$ and the angle $γ$ between these sides. The third side can be determined from the law of cosines:^[4]

c = \sqrt{a^{2} + b^{2} - 2 a b \cos γ} .

Now we use law of cosines to find the second angle:

α = \arccos \frac{b^{2} + c^{2} - a^{2}}{2 b c} .

Finally, $β = 18 0^{\circ} - α - γ .$

File:Resolve triangle with b c beta.png

Two sides and non-included angle given

Two sides and non-included angle given (SSA)

This case is the most difficult and ambiguous. Let two sides $b, c$ and the angle $β$ are known. Equation for the angle $γ$ we can imply from the law of sines:^[5]

\sin γ = \frac{c}{b} \sin β .

We denote further $D = \frac{c}{b} \sin β$ (equation's right side). There are 4 possible cases.

If $D > 1$ , no such triangle exists (the side $b$ «not reaches» to the line BC).
If $D = 1$ , unique solution exists: $γ = 9 0^{\circ}$ , i. e. the triangle is right-angled.

File:Resolve triangle with b c beta 2 solutions.png

Two solutions for triangle

If $D < 1$ two alternatives are possible.
1. If $b < c$ , the angle $γ$ may be acute: $γ = \arcsin D$ or obtuse: $γ^{'} = 18 0^{\circ} - γ$ . The picture on right shows the point $C$ , the side $b$ and the angle $γ$ as the first solution, and the point $C^{'}$ , side $b^{'}$ and the angle $γ^{'}$ as the second solution.
2. If $b ⩾ c$ then $β ⩾ γ$ (the larger side corresponds to a larger angle). Since no triangle can have two obtuse angles, $γ$ is acute angle and the solution $γ = \arcsin D$ is unique.

The third angle $α = 18 0^{\circ} - β - γ$ . The third side can be found from law of sines:

a = b \frac{\sin α}{\sin β}

File:Resolve triangle with c alpha beta.png

A side and two adjacent angles given

A side and two adjacent angles given (ASA)

Known characteristics are the side $c$ and the angles $α, β$ . The third angle $γ = 18 0^{\circ} - α - β .$

Two unknown side can be calculated from the law of sines:^[6]

a = c \frac{\sin α}{\sin γ}; b = c \frac{\sin β}{\sin γ} .

A side, one adjacent angle and the opposite angle given (AAS)

The procedure for solving an AAS triangle is same as that of 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.

File:Spherical triangle 3d opti.png

Spherical triangle

Solving spherical triangles

General form spherical triangle is fully determined by three of its six characteristics (3 sides and 3 angles). Note that the sides of a spherical triangle $a, b, c$ are usually measured rather by angular units than by linear, according to corresponding central angles.

Solution of triangles for non-Euclidean spherical geometry has some differences from the plane case. For example, the sum of the three angles $α + β + γ$ depends on the triangle. In addition, there are no unequal similar triangles, and so the problem of constructing a triangle with specified three angles has a unique solution. Basic relations used to solve a problem are like to the planar case: see Law of cosines (spherical) and Law of sines (spherical).

Among other relationships may be useful half-side formula and Napier's analogies:^[7]

Three sides given

Known: the sides $a, b, c$ (in angular units). Triangle angles are defined from spherical law of cosines:

α = \arccos (\frac{\cos a - \cos b \cos c}{\sin b \sin c}),

β = \arccos (\frac{\cos b - \cos c \cos a}{\sin c \sin a}),

γ = \arccos (\frac{\cos c - \cos a \cos b}{\sin a \sin b}),

Two sides and the included angle given

Known: the sides $a, b$ and the angle $γ$ among it. The side $c$ can be found from the law of cosines:

c = \arccos (\cos a \cos b + \sin a \sin b \cos γ)

The angles $α, β$ can be calculated as above, or by using Napier's analogies:

α = \arctan \frac{2 \sin a}{\tan (\frac{γ}{2}) \sin (b + a) + \cot (\frac{γ}{2}) \sin (b - a)}

β = \arctan \frac{2 \sin b}{\tan (\frac{γ}{2}) \sin (a + b) + \cot (\frac{γ}{2}) \sin (a - b)} .

This problem arises in the navigation problem of finding the great circle between 2 points on the earth specified by their latitude and longitude; in this application, it's 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

\begin{matrix} c & = \arctan \frac{\sqrt{(\sin a \cos b - \cos a \sin b \cos γ)^{2} + (\sin b \sin γ)^{2}}}{\cos a \cos b + \sin a \sin b \cos γ}, \\ α & = \arctan \frac{\sin a \sin γ}{\sin b \cos a - \cos b \sin a \cos γ}, \\ β & = \arctan \frac{\sin b \sin γ}{\sin a \cos b - \cos a \sin b \cos γ}, \end{matrix}

where the signs of the numerators and denominators in these expressions should be used to determine the quadrant of the arctangent.

Two sides and non-included angle given

Known: the sides $b, c$ and the angle $β$ not among it. Solution exists if the following condition takes place:

b > \arcsin (\sin c \sin β)

The angle $γ$ can be found from the Law of sines (spherical):

γ = \arcsin (\frac{\sin c \sin β}{\sin b})

As for the plane case, if $b < c$ then there are two solutions: $γ$ and $18 0^{\circ} - γ$ . Other characteristics we can find by using Napier's analogies:

a = 2 \arctan {\tan (\frac{1}{2} (b - c)) \frac{\sin (\frac{1}{2} (β + γ))}{\sin (\frac{1}{2} (β - γ))}},

α = 2 arccot {\tan (\frac{1}{2} (β - γ)) \frac{\sin (\frac{1}{2} (b + c))}{\sin (\frac{1}{2} (b - c))}} .

A side and two adjacent angles given

Known: the side $c$ and the angles $α, β$ . At first we determine the angle $γ$ using the law of cosines:

γ = \arccos (\sin α \sin β \cos c - \cos α \cos β),

Two unknown sides we can find from the law of cosines (using the calculated angle $γ$ ):

a = \arccos (\frac{\cos α + \cos β \cos γ}{\sin β \sin γ})

b = \arccos (\frac{\cos β + \cos γ \cos α}{\sin γ \sin α})

or by using Napier's analogies:

a = \arctan {\frac{2 \sin α}{\cot (c / 2) \sin (β + α) + \tan (c / 2) \sin (β - α)}},

b = \arctan {\frac{2 \sin β}{\cot (c / 2) \sin (α + β) + \tan (c / 2) \sin (α - β)}},

A side, one adjacent angle and the opposite angle given

Known: the side $a$ and the angles $α, β$ . The side $b$ can be found from the law of sines:

b = \arcsin (\frac{\sin a \sin β}{\sin α}),

If the angle for the side $a$ is acute and $α > β$ , another solution exists:

b = π - \arcsin (\frac{\sin a \sin β}{\sin α})

Other characteristics we can find by using Napier's analogies:

c = 2 \arctan {\tan (\frac{1}{2} (a - b)) \frac{\sin (\frac{1}{2} (α + β))}{\sin (\frac{1}{2} (α - β))}},

γ = 2 arccot {\tan (\frac{1}{2} (α - β)) \frac{\sin (\frac{1}{2} (a + b))}{\sin (\frac{1}{2} (a - b))}},

Three angles given

Known: the angles $α, β, γ$ . From the law of cosines we infer:

a = \arccos (\frac{\cos α + \cos β \cos γ}{\sin β \sin γ}),

b = \arccos (\frac{\cos β + \cos γ \cos α}{\sin γ \sin α}),

c = \arccos (\frac{\cos γ + \cos α \cos β}{\sin α \sin β}) .

Solving right-angled spherical triangles

The above algorithms become much simpler if one of the angles of a triangle (for example, the angle $C$ ) is the right angle. Such a spherical triangle is fully defined by its two elements, and the other three can be calculated using Napier's Pentagon or the following relations.

\sin a = \sin c \cdot \sin A

(from the Law of sines (spherical))

\tan a = \sin b \cdot \tan A

\cos c = \cos a \cdot \cos b

(from the law of cosines (spherical))

\tan b = \tan c \cdot \cos A

\cos A = \cos a \cdot \sin B

(also from the law of cosines)

\cos c = \cot A \cdot \cot B

Some applications

Triangulation

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Suppose you want to measure the distance $d$ from shore to remote ship. You must mark on the shore two points with known distance $l$ between them (base line). Let $α, β$ are the angles between base line and the direction to ship.

From the formulas above (ASA case) one can define the length of the triangle height:

d = \frac{\sin α \sin β}{\sin (α + β)} l = \frac{\tan α \tan β}{\tan α + \tan β} l

This method is used in cabotage. The angles $α, β$ are defined by observations familiar landmarks from the ship.

Another example: you want to measure the height $h$ of a mountain or a high building. The angles $α, β$ from two ground points to the top are specified. Let $l$ be the distance between tis points. From the same ASA case formulas we obtain:

h = \frac{\sin α \sin β}{\sin (β - α)} l = \frac{\tan α \tan β}{\tan β - \tan α} l

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The distance between two points on the globe

That's how to calculate the distance between two points on the globe.

Point A: latitude

λ_{A},

longitude

L_{A}

Point B: latitude

λ_{B},

longitude

L_{B}

We consider the spherical triangle $A B C$ , where $C$ is the north Pole. Some characteristics we know:

a = 9 0^{o} - λ_{B}

b = 9 0^{o} - λ_{A}

γ = L_{A} - L_{B}

It's the case: Two sides and the included angle given. From its formulas we obtain:

A B = R \arccos {\sin λ_{A} \sin λ_{B} + \cos λ_{A} \cos λ_{B} \cos (L_{A} - L_{B})},

Here $R$ is the Earth radius.

References

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My blog: http://www.primaboinca.com/view_profile.php?userid=5889534

External links

Trigonometric Delights, by Eli Maor, Princeton University Press, 1998. Ebook version, in PDF format, full text presented.
Trigonometry by Alfred Monroe Kenyon and Louis Ingold, The Macmillan Company, 1914. In images, full text presented.
Spherical trigonometry on Math World.
Intro to Spherical Trig. Includes discussion of The Napier circle and Napier's rules
Spherical Trigonometry — for the use of colleges and schools by I. Todhunter, M.A., F.R.S. Historical Math Monograph posted by Cornell University Library.
Triangulator – Triangle solver. Solve any triangle problem with the minimum of input data. Drawing of the solved triangle.

de:Dreieck#Berechnung eines beliebigen Dreiecks

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[7] Napier's Analogies at MathWorld

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Iterative refinement: Difference between revisions

en>Monkbot

Revision as of 16:00, 22 January 2014

Contents

Solving plane triangles

Main theorems

Notes

Three sides given (SSS)

Two sides and the included angle given (SAS)

Two sides and non-included angle given (SSA)

A side and two adjacent angles given (ASA)

A side, one adjacent angle and the opposite angle given (AAS)

Solving spherical triangles

Three sides given

Two sides and the included angle given

Two sides and non-included angle given

A side and two adjacent angles given

A side, one adjacent angle and the opposite angle given

Three angles given

Solving right-angled spherical triangles

Some applications

Triangulation

The distance between two points on the globe

See also

References

External links

Navigation menu

@@ Line 1: / Line 1: @@
+{{copy edit|date=June 2013}}
+'''Solution of triangles''' ({{lang-lat|solutio triangulorum}}) is the historical term for solving 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 in the triangle</center>]]
+A general form triangle has 6 main characteristics (see picture): 3 linear (side lengths <math>~a,b,c</math>) and 3 angular (<math>~\alpha,\beta,\gamma</math>). The classical plane trigonometry problem is to specify 3 of the 6 characteristics and  determine the other three. 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.
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+A triangle can be solved 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 | accessdate=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 | accessdate=4 April 2012<!-- 6:42 (UTC)-->}}</ref>
+* Three sides ('''SSS''')
+* Two sides and the included angle ('''SAS''')
+* Two sides and an angle not included between them ('''SSA''')
+* A side and the two angles adjacent to it ('''ASA''')
+* A side, the angle opposite to it and an angle adjacent to it ('''AAS''').
+=== Main theorems ===
+[[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> a^2 = b^2 + c^2 - 2 b c \cos \alpha </math>
+: <math> b^2 = a^2 + c^2 - 2 a c \cos \beta </math>
+: <math> c^2 = a^2 + b^2 - 2 a b \cos \gamma </math>
+; [[Law of sines]]:
+: <math>\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}</math>
+; [[Triangle|Sum of angles]]:
+: <math>\alpha + \beta + \gamma = 180^\circ </math>
+; [[Law of tangents]]
+: <math>\frac{a-b}{a+b} = \frac{\mathrm{tan}[\frac{1}{2}(\alpha-\beta)]}{\mathrm{tan}[\frac{1}{2}(\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, [[law of cosines]] is safer than [[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 \beta = 0.5</math>, the angle <math>\beta</math> can be equal either <math>30^\circ</math> or <math>150^\circ</math>. Using the law of cosines avoids this problem: within the interval from <math>0^\circ</math> to <math>180^\circ</math> the cosine value determines its angle unambiguously.  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 be the solution also. For example, three side lengths uniquely define either a triangle or its reflection.
+[[File:resolve triangle with a b c.png|thumb|right|250px|<center>Three sides given</center>]]
+=== Three sides given (SSS) ===
+Let three side lengths <math>a, b, c</math> be specified. To find the angles <math>\alpha, \beta</math>, you can use the [[law of cosines]]:<ref>{{cite web |url=http://www.mathsisfun.com/algebra/trig-solving-sss-triangles.html |title=Solving SSS Triangles |publisher=Maths is Fun |accessdate=23 Jule 2012 }}</ref>
+: <math> \alpha =  \arccos \frac{b^2 + c^2 - a^2} {2 b c}</math>
+: <math> \beta =  \arccos \frac{a^2 + c^2 - b^2} {2 a c}.</math>
+Then angle <math>\gamma = 180^\circ - \alpha - \beta</math>.
+Some sources recommend to find angle <math>\beta</math> from [[law of sines]] but (as Note 1 above states) there is a risk to confuse acute angle value with obtuse one.
+Another method of calculating the angles from known sides is to apply the [[law of cotangents]].
+[[File:resolve triangle with a b gamma.png|thumb|right|250px|<center>Two sides and the included angle given</center>]]
+=== Two sides and the included angle given (SAS) ===
+Here we know the lengths of sides <math>a, b</math> and the angle <math>\gamma</math> between these sides. 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 |accessdate=24 Jule 2012}}</ref>
+: <math>c = \sqrt{a^2+b^2-2ab\cos\gamma}.</math>
+Now we use law of cosines to find the second angle:
+: <math> \alpha =  \arccos \frac{b^2 + c^2 - a^2} {2 b c}.</math>
+Finally, <math>\beta = 180^\circ - \alpha - \gamma.</math>
+[[File:resolve triangle with b c beta.png|thumb|right|250px|<center>Two sides and non-included angle given</center>]]
+=== Two sides and non-included angle given (SSA) ===
+This case is the most difficult and ambiguous. Let two sides <math>b, c</math> and the angle <math>\beta</math> are known. Equation for the angle <math>\gamma</math> we can imply 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 | accessdate=9 March 2013 }}</ref>
+: <math>\sin\gamma = \frac{c}{b} \sin\beta.</math>
+We denote further <math>~D=\frac{c}{b} \sin\beta</math> (equation's right side). There are 4 possible cases.
+# If <math>D>1</math>, no such triangle exists (the side <math>b</math> «not reaches» to the line BC).
+# If <math>D=1</math>, unique solution exists: <math>\gamma = 90^\circ</math>, i. e. the triangle is [[Right triangle|right-angled]].
+[[File:Resolve triangle with b c beta 2 solutions.png|thumb|right|250px|<center>Two solutions for triangle</center>]]
+# <li value="3"> If <math>D<1</math> two alternatives are possible.
+## If <math>b<c</math>, the angle <math>\gamma</math> may be acute: <math>~\gamma = \arcsin D</math> or obtuse: <math>~\gamma' = 180^\circ - \gamma</math>. The picture on right shows the point <math>C</math>, the side <math>b</math> and the angle <math>\gamma</math> as the first solution, and the point <math>C'</math>, side <math>b'</math> and the angle <math>\gamma'</math> as the second solution.
+## If <math>b \geqslant c</math> then <math>\beta \geqslant \gamma</math> (the larger side corresponds to a larger angle). Since no triangle can have two obtuse angles, <math>~\gamma</math> is acute angle and the solution <math>~\gamma=\arcsin D</math> is unique.
+The third angle <math>\alpha = 180^\circ - \beta - \gamma </math>. The third side can be found from law of sines:
+: <math>a = b\ \frac{\sin\alpha}{\sin\beta}</math>
+[[File:resolve triangle with c alpha beta.png|thumb|right|250px|<center>A side and two adjacent angles given</center>]]
+=== A side and two adjacent angles given (ASA) ===
+Known characteristics are the side  <math>c</math> and the angles <math>\alpha, \beta</math>. The third angle <math>~\gamma = 180^\circ - \alpha - \beta.</math>
+Two unknown side 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 |accessdate=24 Jule 2012}}</ref>
+: <math>a = c\ \frac{\sin\alpha}{\sin\gamma}; \quad b = c\ \frac{\sin\beta}{\sin\gamma}.</math>
+=== A side, one adjacent angle and the opposite angle given (AAS) ===
+The procedure for solving an AAS triangle is same as that of 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]].
+[[File:Spherical triangle 3d opti.png|thumb|250 px|right|<center>Spherical triangle</center>]]
+== Solving spherical triangles ==
+General form [[spherical triangle]] is fully determined by three of its six characteristics (3 sides and 3 angles). Note that the sides of a spherical triangle <math> a, b, c </math> are usually measured rather by angular units than by linear, according to corresponding [[central angle]]s.
+Solution of triangles for [[Non-Euclidean geometry|non-Euclidean]] [[spherical geometry]] has some differences from the plane case. For example, the sum of the three angles <math> \alpha + \beta + \gamma </math> depends on the triangle. In addition, there are no unequal [[Similarity (geometry)|similar triangles]], and so the problem of constructing a triangle with specified three angles has a unique solution. Basic relations used to solve a problem are like to the planar case: see [[Law of cosines (spherical)]] and [[Law of sines#Spherical case|Law of sines (spherical)]].
+Among other relationships may be useful [[half-side formula]] and [[Napier's analogies]]:<ref>[http://mathworld.wolfram.com/NapiersAnalogies.html Napier's Analogies] at MathWorld</ref>
+* <math>\tan\frac{c}{2} \cos\frac{\alpha-\beta}{2} = \tan\frac{a+b}{2} \cos\frac{\alpha+\beta}{2}</math>
+* <math>\tan\frac{c}{2} \sin\frac{\alpha-\beta}{2} = \tan\frac{a-b}{2} \sin\frac{\alpha+\beta}{2}</math>
+* <math>\cot\frac{\gamma}{2} \cos\frac{a-b}{2} = \tan\frac{\alpha+\beta}{2} \cos\frac{a+b}{2}</math>
+* <math>\cot\frac{\gamma}{2} \sin\frac{a-b}{2} = \tan\frac{\alpha-\beta}{2} \sin\frac{a+b}{2}.</math>
+[[File:Solve spherical triangle with a b c.png|thumb|right|250px|<center>Three sides given</center>]]
+=== Three sides given ===
+Known: the sides <math>a, b, c</math> (in angular units). Triangle angles are defined from [[law of cosines (spherical)|spherical law of cosines]]:
+: <math>\alpha = \arccos\left(\frac{\cos a-\cos b\ \cos c}{\sin b\ \sin c}\right),</math>
+: <math>\beta  = \arccos\left(\frac{\cos b-\cos c\ \cos a}{\sin c\ \sin a}\right),</math>
+: <math>\gamma = \arccos\left(\frac{\cos c-\cos a\ \cos b}{\sin a\ \sin b}\right),</math>
+<br clear="all" />
+[[File:Solve spherical triangle with a b gamma.png|thumb|right|250px|<center>Two sides and the included angle given</center>]]
+=== Two sides and the included angle given ===
+Known: the sides <math>a, b</math> and the angle <math>\gamma</math> among it. The side <math>c</math> can be found from the law of cosines:
+: <math>c = \arccos \left(\cos a\cos b + \sin a\sin b\cos\gamma \right)</math>
+The angles <math>\alpha, \beta</math> can be calculated as above, or by using Napier's analogies:
+: <math>\alpha = \arctan\ \frac{2\sin a}{\tan(\frac{\gamma}{2}) \sin (b+a) + \cot(\frac{\gamma}{2})\sin(b-a)}</math>
+: <math>\beta = \arctan\ \frac{2\sin b}{\tan(\frac{\gamma}{2}) \sin (a+b) + \cot(\frac{\gamma}{2})\sin(a-b) }.</math>
+This problem arises in the [[Great-circle navigation#Course and distance|navigation problem]]
+of finding the great circle between 2 points on the earth specified by their
+latitude and longitude; in this application, it's 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>\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},\\
+\alpha &= \arctan\frac
+{\sin a\sin\gamma}
+{\sin b\cos a - \cos b\sin a\cos\gamma},\\
+\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.
+<br clear="all" />
+[[File:Solve spherical triangle with b c beta.png|thumb|right|250px|<center>Two sides and non-included angle given</center>]]
+=== Two sides and non-included angle given ===
+Known: the sides <math>b, c</math> and the angle <math>\beta</math> not among it. Solution exists if the following condition takes place:
+: <math>b > \arcsin (\sin c\,\sin\beta)</math>
+The angle <math>\gamma</math> can be found from the [[Law of sines#Spherical case|Law of sines (spherical)]]:
+: <math>\gamma = \arcsin \left(\frac{\sin c\,\sin\beta}{\sin b}\right)</math>
+As for the plane case, if <math>b<c</math> then there are two solutions: <math>\gamma</math> and <math>~180^\circ - \gamma</math>.
+Other characteristics we can find by using Napier's analogies:
+: <math>a = 2\arctan \left\{ \tan\left(\frac12(b-c)\right) \frac{\sin \left(\frac12(\beta+\gamma)\right)}{\sin\left(\frac12(\beta-\gamma)\right)} \right\},</math>
+: <math>\alpha = 2\arccot \left\{\tan\left(\frac12(\beta-\gamma)\right) \frac{\sin \left(\frac12(b+c)\right)}{\sin \left(\frac12(b-c)\right)} \right\}.</math>
+[[File:Solve spherical triangle with c alpha beta.png|thumb|right|250px|<center>A side and two adjacent angles given</center>]]
+=== A side and two adjacent angles given ===
+Known: the side <math>c</math> and the angles <math>\alpha, \beta</math>. At first we determine the angle <math>\gamma</math> using the [[Law of cosines (spherical)|law of cosines]]:
+: <math>\gamma = \arccos(\sin\alpha\sin\beta\cos c -\cos\alpha\cos\beta),\,</math>
+Two unknown sides we can find from the law of cosines (using the calculated angle <math>\gamma</math>):
+: <math>a=\arccos\left(\frac{\cos\alpha+\cos\beta\cos\gamma}{\sin\beta\sin\gamma}\right)</math>
+: <math>b=\arccos\left(\frac{\cos\beta+\cos\gamma\cos\alpha}{\sin\gamma\sin\alpha}\right)</math>
+or by using Napier's analogies:
+: <math>a = \arctan\left\{\frac{2\sin\alpha}{\cot(c/2) \sin(\beta+\alpha) + \tan(c/2) \sin(\beta-\alpha)}\right\},</math>
+: <math>b = \arctan\left\{\frac{2\sin\beta} {\cot(c/2) \sin(\alpha+\beta) + \tan(c/2)\sin(\alpha-\beta)}\right\},</math>
+<br clear="all" />
+[[File:Solve spherical triangle with a alpha beta.png|thumb|right|250px|<center>A side, one adjacent angle and the opposite angle given</center>]]
+=== A side, one adjacent angle and the opposite angle given ===
+Known: the side <math>a</math> and the angles <math>\alpha, \beta</math>. The side <math>b</math> can be found from the [[Law of sines#Spherical case|law of sines]]:
+: <math>b = \arcsin \left( \frac{\sin a\,\sin \beta}{\sin \alpha} \right),</math>
+If the angle for the side <math>a</math> is acute and <math>\alpha > \beta</math>, another solution exists:
+: <math>b = \pi - \arcsin \left( \frac{\sin a\,\sin \beta}{\sin \alpha} \right)</math>
+Other characteristics we can find by using Napier's analogies:
+: <math>c =  2\arctan \left\{ \tan\left(\frac12(a-b)\right) \frac{\sin\left(\frac12(\alpha+\beta)\right)}{\sin\left(\frac12(\alpha-\beta)\right)}\right\},</math>
+: <math>\gamma = 2\arccot \left\{\tan\left(\frac12(\alpha-\beta)\right) \frac{\sin \left(\frac12(a+b)\right)}{\sin \left(\frac12(a-b)\right)} \right\},</math>
+<br clear="all" />
+[[File:Solve spherical triangle with alpha beta gamma.png|thumb|right|250px|<center>Three angles given</center>]]
+=== Three angles given ===
+Known: the angles <math>\alpha, \beta, \gamma</math>. From the [[Law of cosines (spherical)|law of cosines]] we infer:
+: <math>a=\arccos\left(\frac{\cos\alpha+\cos\beta\cos\gamma}{\sin\beta\sin\gamma}\right),</math>
+: <math>b=\arccos\left(\frac{\cos\beta+\cos\gamma\cos\alpha}{\sin\gamma\sin\alpha}\right),</math>
+: <math>c=\arccos\left(\frac{\cos\gamma+\cos\alpha\cos\beta}{\sin\alpha\sin\beta}\right).</math>
+<br clear="all" />
+=== Solving right-angled spherical triangles ===
+The above algorithms become much simpler if one of the angles of a triangle (for example, the angle <math>C</math>) 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 [[Law of sines#Spherical case|Law of sines (spherical)]])
+: <math>\tan a = \sin b \cdot \tan A</math>
+: <math>\cos c = \cos a \cdot \cos b</math> (from the [[law of cosines (spherical)]])
+: <math>\tan b = \tan c \cdot \cos A</math>
+: <math>\cos A = \cos a \cdot \sin B</math> (also from the 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]]</center>]]
+  {{main|Triangulation}}
+Suppose you want to measure the distance <math>d</math>  from shore to remote ship. You must mark on the shore two points with known distance <math>l</math> between them (base line). Let <math>\alpha,\beta</math> are the angles between base line and the direction to ship.
+From the formulas above (ASA case) one can define the length of the [[Triangle|triangle height]]:
+: <math>d = \frac{\sin\alpha\,\sin\beta}{\sin(\alpha+\beta)} \,l = \frac{\tan\alpha\,\tan\beta}{\tan\alpha+\tan\beta} \,l</math>
+This method is used in [[cabotage]]. The angles <math>\alpha, \beta</math> are defined  by observations familiar landmarks from the ship.
+<br clear=all>
+[[File:mountain height by triangulation.png|thumb|right|320px|<center>How to measure a mountain height</center>]]
+Another example: you want to measure the height <math>h</math> of a mountain or a high building. The angles <math>\alpha, \beta</math> from two ground points to the top are specified. Let <math>l</math> be the distance between tis points. From the same ASA case formulas we obtain:
+: <math> h = \frac{\sin\alpha\,\sin\beta}{\sin(\beta-\alpha)} \,l = \frac{\tan\alpha\,\tan\beta}{\tan\beta-\tan\alpha} \,l</math>
+{{clr}}
+=== The distance between two points on the globe ===
+[[File:distance on earth.png|224px|right]]
+That's how to calculate the distance between two points on the globe.
+: Point A: latitude  <math>\lambda_\mathrm{A},</math> longitude  <math>L_\mathrm{A}</math>
+: Point B: latitude <math>\lambda_\mathrm{B},</math> longitude  <math>L_\mathrm{B}</math>
+We consider the spherical triangle<math>ABC</math>, where <math>C</math> is the north Pole. Some characteristics we know:
+: <math>a = 90^\mathrm{o} - \lambda_\mathrm{B}\,</math>
+: <math>b = 90^\mathrm{o} - \lambda_\mathrm{A}\,</math>
+: <math>\gamma = L_\mathrm{A}-L_\mathrm{B}\,</math>
+It's the case: [[#Two sides and the included angle given|Two sides and the included angle given]]. From its formulas we obtain:
+: <math>\mathrm{AB} = R \arccos\left\{\sin \lambda_\mathrm{A} \,\sin \lambda_\mathrm{B} + \cos \lambda_\mathrm{A} \,\cos \lambda_\mathrm{B} \,\cos \left(L_\mathrm{A}-L_\mathrm{B}\right)\right\},</math>
+Here <math>R</math> is the [[Earth|Earth radius]].
+== See also ==
+* [[Congruence (geometry)|Congruence]]
+* [[Hansen's problem]]
+* [[Hinge theorem]]
+* [[Snellius–Pothenot problem]]
+* [[Lenart Sphere]]
+== References ==
+{{reflist}}
+* {{cite book | author=Euclid | authorlink=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 | origyear=1925 | isbn=0-486-60088-2 }}
+== External links ==
+* [http://www.pupress.princeton.edu/books/maor/ Trigonometric Delights], by [[Eli Maor]], Princeton University Press, 1998.  Ebook version, in PDF format, full text presented.
+* [http://baqaqi.chi.il.us/buecher/mathematics/trigonometry/index.html Trigonometry] by Alfred Monroe Kenyon and Louis Ingold, The Macmillan Company, 1914. In images, full text presented.
+* ''[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
+* [http://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 triangle problem with the minimum of input data. Drawing of the solved triangle.
+[[Category:Trigonometry]]
+[[Category:Spherical trigonometry]]
+[[Category:Triangle geometry]]
+[[de:Dreieck#Berechnung eines beliebigen Dreiecks]]

Iterative refinement: Difference between revisions

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Solving plane triangles

Main theorems

Notes

Three sides given (SSS)

Two sides and the included angle given (SAS)

Two sides and non-included angle given (SSA)

A side and two adjacent angles given (ASA)

A side, one adjacent angle and the opposite angle given (AAS)

Solving spherical triangles

Three sides given

Two sides and the included angle given

Two sides and non-included angle given

A side and two adjacent angles given

A side, one adjacent angle and the opposite angle given

Three angles given

Solving right-angled spherical triangles

Some applications

Triangulation

The distance between two points on the globe

See also

References

External links

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