# Solution of triangles

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Solution of triangles (Latin: solutio triangulorum) is the historical term for the solving of the main trigonometric problem: to find the characteristics of the triangle (three angles, the lengths of the three sides etc) when some (but not all) of this characteristics are given. The triangle can be located on a plane or on a sphere. This problem often occurs in various trigonometric applications, such as geodesy, astronomy, construction, navigation etc.

## Solving plane triangles

Standard notation in the triangle

In a general form triangle, there are 6 main characteristics (see picture): 3 linear (side lengths $~a,b,c$) and 3 angular ($~\alpha,\beta,\gamma$). The classical plane trigonometry problem is to specify 3 of the 6 characteristics and to determine the three others. Obviously, if we know only 2 or 3 angles, solution is undefined, because any triangle similar to a solution is the solution also, so we assume that at least one of the known values is linear.

Thus a triangle can be solved when given the any of the following information:[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

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

1. 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 $\sin \beta = 0.5$, 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 determines its angle unambiguously.
2. 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.
Three sides given

### Three sides given (SSS)

Let three side lengths $a, b, c$ are specified. To find the angles $\alpha, \beta$, you can use the law of cosines:[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 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.

Two sides and the included angle given

### Two sides and the included angle given (SAS)

Let we know the length of sides $a, b$ and the angle $\gamma$ between this sides. 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)

This case is the most difficult and ambiguous. Let two sides $b, c$ and the angle $\beta$ are known. Equation for the angle $\gamma$ we can imply 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 4 possible cases.

1. If $D>1$, no such triangle exists (the side $b$ «not reaches» to the line BC).
2. If $D=1$, 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, 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.

The third angle $\alpha = 180^\circ - \beta - \gamma$. The third side can be found from 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)

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)

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.

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 $\alpha + \beta + \gamma$ 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

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

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:

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

Two sides and non-included angle given

### Two sides and non-included angle given

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 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

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

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

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

The above algorithms become much simpler if one of the angles of a triangle (for example, the angle $C$) is the right angle. Such spherical triangle is fully defined by its two elements, 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

Distance measurement by 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

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 triangle$ABC$, 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.

## References

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.
4. ^ "Solving SAS Triangles". Maths is Fun. Retrieved 24 Jule 2012.
5. ^ "Solving SSA Triangles". Maths is Fun. Retrieved 9 March 2013.
6. ^ "Solving ASA Triangles". Maths is Fun. Retrieved 24 Jule 2012.
7. ^ Napier's Analogies at MathWorld