Solution of triangles

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Solution of triangles (Latin: solutio triangulorum) 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[edit]

Standard notation in the triangle

A general form triangle has six main characteristics (see picture): three linear (side lengths ~a,b,c) and three angular (~\alpha,\beta,\gamma). The classical plane trigonometry problem is to specify three of the six 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[edit]

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 \alpha
 b^2 = a^2 + c^2 - 2 a c \cos \beta
 c^2 = a^2 + b^2 - 2 a b \cos \gamma
Law of sines
\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma}
Sum of angles
\alpha + \beta + \gamma = 180^\circ
Law of tangents
\frac{a-b}{a+b} = \frac{\mathrm{tan}[\frac{1}{2}(\alpha-\beta)]}{\mathrm{tan}[\frac{1}{2}(\alpha+\beta)]}.

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

Notes[edit]

  1. 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 \sin \beta = 0.5, the angle \beta can be equal either 30^\circ or 150^\circ. Using the law of cosines avoids this problem: within the interval from 0^\circ to 180^\circ 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).
  2. 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

Three sides given (SSS)[edit]

Let three side lengths a, b, c be specified. To find the angles \alpha, \beta, the law of cosines can be used:[3]

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

Then angle \gamma = 180^\circ - \alpha - \beta.

Some sources recommend to find angle \beta 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.

Two sides and the included angle given

Two sides and the included angle given (SAS)[edit]

Here the lengths of sides a, b and the angle \gamma between these sides are known. The third side can be determined from the law of cosines:[4]

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

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

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

Finally, \beta = 180^\circ - \alpha - \gamma.

Two sides and non-included angle given

Two sides and non-included angle given (SSA)[edit]

This case is the most difficult and ambiguous. Assume that two sides b, c and the angle \beta are known. The equation for the angle \gamma can be implied from the law of sines:[5]

\sin\gamma = \frac{c}{b} \sin\beta.

We denote further ~D=\frac{c}{b} \sin\beta (equation's right side). There are four possible cases:

  1. If D>1, no such triangle exists because the side b does not reach line BC. For the same reason a solution does not exist if the angle \beta \geqslant 90^\circ and b \leqslant c.
  2. If D=1, a unique solution exists: \gamma = 90^\circ, i.e., the triangle is right-angled.
Two solutions for triangle
  1. If D<1 two alternatives are possible.
    1. If b<c, the angle \gamma may be acute: ~\gamma = \arcsin D or obtuse: ~\gamma' = 180^\circ - \gamma. The picture on right shows the point C, the side b and the angle \gamma as the first solution, and the point C', side b' and the angle \gamma' as the second solution.
    2. If b \geqslant c then \beta \geqslant \gamma (the larger side corresponds to a larger angle). Since no triangle can have two obtuse angles, ~\gamma is acute angle and the solution ~\gamma=\arcsin D is unique.

Once \gamma is obtained, the third angle \alpha = 180^\circ - \beta - \gamma .

The third side can then be found from the law of sines:

a = b\ \frac{\sin\alpha}{\sin\beta}
A side and two adjacent angles given

A side and two adjacent angles given (ASA)[edit]

The known characteristics are the side c and the angles \alpha, \beta. The third angle ~\gamma = 180^\circ - \alpha - \beta.

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

a = c\ \frac{\sin\alpha}{\sin\gamma}; \quad b = c\ \frac{\sin\beta}{\sin\gamma}.

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

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.

Spherical triangle

Solving spherical triangles[edit]

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  \alpha + \beta + \gamma 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]

  • \tan\frac{c}{2} \cos\frac{\alpha-\beta}{2} = \tan\frac{a+b}{2} \cos\frac{\alpha+\beta}{2}
  • \tan\frac{c}{2} \sin\frac{\alpha-\beta}{2} = \tan\frac{a-b}{2} \sin\frac{\alpha+\beta}{2}
  • \cot\frac{\gamma}{2} \cos\frac{a-b}{2} = \tan\frac{\alpha+\beta}{2} \cos\frac{a+b}{2}
  • \cot\frac{\gamma}{2} \sin\frac{a-b}{2} = \tan\frac{\alpha-\beta}{2} \sin\frac{a+b}{2}.
Three sides given

Three sides given[edit]

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

\alpha = \arccos\left(\frac{\cos a-\cos b\ \cos c}{\sin b\ \sin c}\right),
\beta  = \arccos\left(\frac{\cos b-\cos c\ \cos a}{\sin c\ \sin a}\right),
\gamma = \arccos\left(\frac{\cos c-\cos a\ \cos b}{\sin a\ \sin b}\right),


Two sides and the included angle given

Two sides and the included angle given[edit]

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

c = \arccos \left(\cos a\cos b + \sin a\sin b\cos\gamma \right)

The angles \alpha, \beta can be calculated as above, or by using Napier's analogies:

\alpha = \arctan\ \frac{2\sin a}{\tan(\frac{\gamma}{2}) \sin (b+a) + \cot(\frac{\gamma}{2})\sin(b-a)}
\beta = \arctan\ \frac{2\sin b}{\tan(\frac{\gamma}{2}) \sin (a+b) + \cot(\frac{\gamma}{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{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}

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

Two sides and non-included angle given[edit]

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

b > \arcsin (\sin c\,\sin\beta)

The angle \gamma can be found from the Law of sines (spherical):

\gamma = \arcsin \left(\frac{\sin c\,\sin\beta}{\sin b}\right)

As for the plane case, if b<c then there are two solutions: \gamma and ~180^\circ - \gamma. Other characteristics we can find by using Napier's analogies:

a = 2\arctan \left\{ \tan\left(\frac12(b-c)\right) \frac{\sin \left(\frac12(\beta+\gamma)\right)}{\sin\left(\frac12(\beta-\gamma)\right)} \right\},
\alpha = 2\arccot \left\{\tan\left(\frac12(\beta-\gamma)\right) \frac{\sin \left(\frac12(b+c)\right)}{\sin \left(\frac12(b-c)\right)} \right\}.
A side and two adjacent angles given

A side and two adjacent angles given[edit]

Known: the side c and the angles \alpha, \beta. At first we determine the angle \gamma using the law of cosines:

\gamma = \arccos(\sin\alpha\sin\beta\cos c -\cos\alpha\cos\beta),\,

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

a=\arccos\left(\frac{\cos\alpha+\cos\beta\cos\gamma}{\sin\beta\sin\gamma}\right)
b=\arccos\left(\frac{\cos\beta+\cos\gamma\cos\alpha}{\sin\gamma\sin\alpha}\right)

or by using Napier's analogies:

a = \arctan\left\{\frac{2\sin\alpha}{\cot(c/2) \sin(\beta+\alpha) + \tan(c/2) \sin(\beta-\alpha)}\right\},
b = \arctan\left\{\frac{2\sin\beta} {\cot(c/2) \sin(\alpha+\beta) + \tan(c/2)\sin(\alpha-\beta)}\right\},


A side, one adjacent angle and the opposite angle given

A side, one adjacent angle and the opposite angle given[edit]

Known: the side a and the angles \alpha, \beta. The side b can be found from the law of sines:

b = \arcsin \left( \frac{\sin a\,\sin \beta}{\sin \alpha} \right),

If the angle for the side a is acute and \alpha > \beta, another solution exists:

b = \pi - \arcsin \left( \frac{\sin a\,\sin \beta}{\sin \alpha} \right)

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

c =  2\arctan \left\{ \tan\left(\frac12(a-b)\right) \frac{\sin\left(\frac12(\alpha+\beta)\right)}{\sin\left(\frac12(\alpha-\beta)\right)}\right\},
\gamma = 2\arccot \left\{\tan\left(\frac12(\alpha-\beta)\right) \frac{\sin \left(\frac12(a+b)\right)}{\sin \left(\frac12(a-b)\right)} \right\},


Three angles given

Three angles given[edit]

Known: the angles \alpha, \beta, \gamma. From the law of cosines we infer:

a=\arccos\left(\frac{\cos\alpha+\cos\beta\cos\gamma}{\sin\beta\sin\gamma}\right),
b=\arccos\left(\frac{\cos\beta+\cos\gamma\cos\alpha}{\sin\gamma\sin\alpha}\right),
c=\arccos\left(\frac{\cos\gamma+\cos\alpha\cos\beta}{\sin\alpha\sin\beta}\right).


Solving right-angled spherical triangles[edit]

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[edit]

Triangulation[edit]

Distance measurement by triangulation
Main article: Triangulation

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 \alpha,\beta 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\alpha\,\sin\beta}{\sin(\alpha+\beta)} \,l = \frac{\tan\alpha\,\tan\beta}{\tan\alpha+\tan\beta} \,l

This method is used in cabotage. The angles \alpha, \beta are defined by observations familiar landmarks from the ship.

How to measure a mountain height

Another example: you want to measure the height h of a mountain or a high building. The angles \alpha, \beta 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\alpha\,\sin\beta}{\sin(\beta-\alpha)} \,l = \frac{\tan\alpha\,\tan\beta}{\tan\beta-\tan\alpha} \,l


The distance between two points on the globe[edit]

Distance on earth.png

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

Point A: latitude \lambda_\mathrm{A}, longitude L_\mathrm{A}
Point B: latitude \lambda_\mathrm{B}, longitude L_\mathrm{B}

We consider the spherical triangleABC, where C is the north Pole. Some characteristics we know:

a = 90^\mathrm{o} - \lambda_\mathrm{B}\,
b = 90^\mathrm{o} - \lambda_\mathrm{A}\,
\gamma = L_\mathrm{A}-L_\mathrm{B}\,

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

\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\},

Here R is the Earth radius.

See also[edit]

References[edit]

  1. ^ "Solving Triangles". Maths is Fun. Retrieved 4 April 2012. 
  2. ^ "Solving Triangles". web.horacemann.org. Retrieved 4 April 2012. 
  3. ^ "Solving SSS Triangles". Maths is Fun. Retrieved 23 Jule 2012.  Check date values in: |accessdate= (help)
  4. ^ "Solving SAS Triangles". Maths is Fun. Retrieved 24 Jule 2012.  Check date values in: |accessdate= (help)
  5. ^ "Solving SSA Triangles". Maths is Fun. Retrieved 9 March 2013. 
  6. ^ "Solving ASA Triangles". Maths is Fun. Retrieved 24 Jule 2012.  Check date values in: |accessdate= (help)
  7. ^ Napier's Analogies at MathWorld

External links[edit]