Spherical trigonometry

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The octant of a sphere is a spherical triangle with three right angles.

Spherical trigonometry is the branch of spherical geometry that deals with the metrical relationships between the sides and angles of spherical triangles, traditionally expressed using trigonometric functions. On the sphere, geodesics are great circles. Spherical trigonometry is of great importance for calculations in astronomy, geodesy, and navigation.

The origins of spherical trigonometry in Greek mathematics and the major developments in Islamic mathematics are discussed fully in History of trigonometry and Mathematics in medieval Islam. The subject came to fruition in Early Modern times with important developments by John Napier, Delambre and others, and attained an essentially complete form by the end of the nineteenth century with the publication of Todhunter's textbook Spherical trigonometry for the use of colleges and Schools. Since then, significant developments have been the application of vector methods, quaternion methods, and the use of numerical methods.

Preliminaries

Eight spherical triangles defined by the intersection of three great circles.

Spherical polygons

A spherical polygon is a polygon on the surface of the sphere. Its sides are arcs of great circles—the spherical geometry equivalent of line segments in plane geometry.

Such polygons may have any number of sides greater than 1. Two-sided spherical polygons—lunes, also called digons or bi-angles—are bounded by two great-circle arcs: a familiar example is the curved outward-facing surface of a segment of an orange. Three arcs serve to define a spherical triangle, the principal subject of this article. Polygons with higher numbers of sides (4-sided spherical quadrilaterals, 5-sided spherical pentagons, etc.) are defined in similar manner. Analogously to their plane counterparts, spherical polygons with more than 3 sides can always be treated as the composition of spherical triangles.

One spherical polygon with interesting properties is the pentagramma mirificum, a 5-sided spherical star polygon with a right angle at every vertex.

From this point in the article, discussion will be restricted to spherical triangles, referred to simply as triangles.

Notation

The basic triangle on a unit sphere.

• Both vertices and angles at the vertices of a triangle are denoted by the same upper case letters A, B, and C.
• Sides are denoted by lower-case letters: a, b, and c.
• The angle A (respectively, B and C) may be regarded either as the angle between the two planes that intersect the sphere at the vertex A, or, equivalently, as the angle between the tangents of the great circle arcs where they meet at the vertex.
• Angles are expressed in radians. The angles of proper spherical triangles are (by convention) less than π, so that
  $${\displaystyle \pi <A+B+C<3\pi }$$

  .

In particular, the sum of the angles of a spherical triangle is strictly greater than the sum of the angles of a triangle defined on the Euclidean plane, which is always exactly π radians.

• Sides are also expressed in radians. A side (regarded as a great circle arc) is measured by the angle that it subtends at the centre. On the unit sphere, this radian measure is numerically equal to the arc length. By convention, the sides of proper spherical triangles are less than π, so that
  $${\displaystyle 0<a+b+c<2\pi }$$

  .
• The sphere's radius is taken as unity. For specific practical problems on a sphere of radius R the measured lengths of the sides must be divided by R before using the identities given below. Likewise, after a calculation on the unit sphere the sides a, b, and c must be multiplied by R.

Polar triangles

The polar triangle △A'B'C'

The polar triangle associated with a triangle △ABC is defined as follows. Consider the great circle that contains the side BC. This great circle is defined by the intersection of a diametral plane with the surface. Draw the normal to that plane at the centre: it intersects the surface at two points and the point that is on the same side of the plane as A is (conventionally) termed the pole of A and it is denoted by A'. The points B' and C' are defined similarly.

The triangle △A'B'C' is the polar triangle corresponding to triangle △ABC. A very important theorem proves that the angles and sides of the polar triangle are given by

$${\displaystyle \begin{array}{rlrlrl}{A}^{\prime }& =\pi -a,& B^{\prime }& =\pi -b,& C^{\prime }& =\pi -c,\\ a^{\prime }& =\pi -A,& b^{\prime }& =\pi -B,& c^{\prime }& =\pi -C.\end{array}}$$

Therefore, if any identity is proved for △ABC then we can immediately derive a second identity by applying the first identity to the polar triangle by making the above substitutions. This is how the supplemental cosine equations are derived from the cosine equations. Similarly, the identities for a quadrantal triangle can be derived from those for a right-angled triangle. The polar triangle of a polar triangle is the original triangle.

Cosine rules and sine rules

Cosine rules

Main article: Spherical law of cosines

The cosine rule is the fundamental identity of spherical trigonometry: all other identities, including the sine rule, may be derived from the cosine rule:

$${\displaystyle \begin{array}{rl}\cos a& =\cos b\cos c+\sin b\sin c\cos A,\\ \cos b& =\cos c\cos a+\sin c\sin a\cos B,\\ \cos c& =\cos a\cos b+\sin a\sin b\cos C.\end{array}}$$

These identities generalize the cosine rule of plane trigonometry, to which they are asymptotically equivalent in the limit of small interior angles. (On the unit sphere, if ${\displaystyle a,b,c\to 0}$ set ${\displaystyle \sin a\approx a}$ and ${\displaystyle (\cos a-\cos b{)}^2\approx 0}$ etc.; see Spherical law of cosines.)

Sine rules

Main article: Spherical law of sines

The spherical law of sines is given by the formula

$${\displaystyle \frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c}.}$$

These identities approximate the sine rule of plane trigonometry when the sides are much smaller than the radius of the sphere.

Derivation of the cosine rule

Main article: Spherical law of cosines

The spherical cosine formulae were originally proved by elementary geometry and the planar cosine rule (Todhunter, Art.37). He also gives a derivation using simple coordinate geometry and the planar cosine rule (Art.60). The approach outlined here uses simpler vector methods. (These methods are also discussed at Spherical law of cosines.)

Consider three unit vectors OA→, OB→, OC→ drawn from the origin to the vertices of the triangle (on the unit sphere). The arc BC subtends an angle of magnitude a at the centre and therefore OB→ · OC→ = cos a. Introduce a Cartesian basis with OA→ along the z-axis and OB→ in the xz-plane making an angle c with the z-axis. The vector OC→ projects to ON in the xy-plane and the angle between ON and the x-axis is A. Therefore, the three vectors have components:

$${\displaystyle \begin{array}{rl}\overrightarrow{OA}:& (0,0,1)\\ \overrightarrow{OB}:& (\sin c,0,\cos c)\\ \overrightarrow{OC}:& (\sin b\cos A,\sin b\sin A,\cos b).\end{array}}$$

The scalar product OB→ · OC→ in terms of the components is

$${\displaystyle \overrightarrow{OB}\cdot \overrightarrow{OC}=\sin c\sin b\cos A+\cos c\cos b.}$$

Equating the two expressions for the scalar product gives

$${\displaystyle \cos a=\cos b\cos c+\sin b\sin c\cos A.}$$

This equation can be re-arranged to give explicit expressions for the angle in terms of the sides:

$${\displaystyle \cos A=\frac{\cos a-\cos b\cos c}{\sin b\sin c}.}$$

The other cosine rules are obtained by cyclic permutations.

Derivation of the sine rule

Main article: Spherical law of sines

This derivation is given in Todhunter, (Art.40). From the identity ${\displaystyle {\sin }^{2}A=1-{\cos }^{2}A}$ and the explicit expression for cos A given immediately above

$${\displaystyle \begin{array}{rl}{\sin }^{2}A& =1-{\left(\frac{\cos a-\cos b\cos c}{\sin b\sin c}\right)}^2\\ & =\frac{(1-{\cos }^{2}b)(1-{\cos }^{2}c)-(\cos a-\cos b\cos c{)}^2}{{\sin }^{2}b{\sin }^{2}c}\\ \frac{\sin A}{\sin a}& =\frac{\sqrt{1-{\cos }^{2}a-{\cos }^{2}b-{\cos }^{2}c+2\cos a\cos b\cos c}}{\sin a\sin b\sin c}.\end{array}}$$

Since the right hand side is invariant under a cyclic permutation of a, b, and c the spherical sine rule follows immediately.

Alternative derivations

There are many ways of deriving the fundamental cosine and sine rules and the other rules developed in the following sections. For example, Todhunter gives two proofs of the cosine rule (Articles 37 and 60) and two proofs of the sine rule (Articles 40 and 42). The page on Spherical law of cosines gives four different proofs of the cosine rule. Text books on geodesy and spherical astronomy give different proofs and the online resources of MathWorld provide yet more. There are even more exotic derivations, such as that of Banerjee who derives the formulae using the linear algebra of projection matrices and also quotes methods in differential geometry and the group theory of rotations.

The derivation of the cosine rule presented above has the merits of simplicity and directness and the derivation of the sine rule emphasises the fact that no separate proof is required other than the cosine rule. However, the above geometry may be used to give an independent proof of the sine rule. The scalar triple product, OA→ · (OB→ × OC→) evaluates to sin b sin c sin A in the basis shown. Similarly, in a basis oriented with the z-axis along OB→, the triple product OB→ · (OC→ × OA→), evaluates to sin c sin a sin B. Therefore, the invariance of the triple product under cyclic permutations gives sin b sin A = sin a sin B which is the first of the sine rules. See curved variations of the law of sines to see details of this derivation.

Identities

Supplemental cosine rules

Applying the cosine rules to the polar triangle gives (Todhunter, Art.47), i.e. replacing A by π – a, a by π – A etc.,

$${\displaystyle \begin{array}{rl}\cos A& =-\cos B\cos C+\sin B\sin C\cos a,\\ \cos B& =-\cos C\cos A+\sin C\sin A\cos b,\\ \cos C& =-\cos A\cos B+\sin A\sin B\cos c.\end{array}}$$

Cotangent four-part formulae

The six parts of a triangle may be written in cyclic order as (aCbAcB). The cotangent, or four-part, formulae relate two sides and two angles forming four consecutive parts around the triangle, for example (aCbA) or BaCb). In such a set there are inner and outer parts: for example in the set (BaCb) the inner angle is C, the inner side is a, the outer angle is B, the outer side is b. The cotangent rule may be written as (Todhunter, Art.44)

$${\displaystyle \cos (\begin{array}{c}\text{inner}\\ \text{side}\end{array})\cos (\begin{array}{c}\text{inner}\\ \text{angle}\end{array})=\cot (\begin{array}{c}\text{outer}\\ \text{side}\end{array})\sin (\begin{array}{c}\text{inner}\\ \text{side}\end{array})-\cot (\begin{array}{c}\text{outer}\\ \text{angle}\end{array})\sin (\begin{array}{c}\text{inner}\\ \text{angle}\end{array}),}$$

and the six possible equations are (with the relevant set shown at right):

$${\displaystyle \begin{array}{rlrlrl}\text{(CT1)}& & \cos b\cos C& =\cot a\sin b-\cot A\sin C& & (aCbA)\\ \text{(CT2)}& & \cos b\cos A& =\cot c\sin b-\cot C\sin A& & (CbAc)\\ \text{(CT3)}& & \cos c\cos A& =\cot b\sin c-\cot B\sin A& & (bAcB)\\ \text{(CT4)}& & \cos c\cos B& =\cot a\sin c-\cot A\sin B& & (AcBa)\\ \text{(CT5)}& & \cos a\cos B& =\cot c\sin a-\cot C\sin B& & (cBaC)\\ \text{(CT6)}& & \cos a\cos C& =\cot b\sin a-\cot B\sin C& & (BaCb)\end{array}}$$

To prove the first formula start from the first cosine rule and on the right-hand side substitute for cos c from the third cosine rule:

$${\displaystyle \begin{array}{rl}\cos a& =\cos b\cos c+\sin b\sin c\cos A\\ & =\cos b(\cos a\cos b+\sin a\sin b\cos C)+\sin b\sin C\sin a\cot A\\ \cos a{\sin }^{2}b& =\cos b\sin a\sin b\cos C+\sin b\sin C\sin a\cot A.\end{array}}$$

The result follows on dividing by sin a sin b. Similar techniques with the other two cosine rules give CT3 and CT5. The other three equations follow by applying rules 1, 3 and 5 to the polar triangle.

Half-angle and half-side formulae

With ${\displaystyle 2s=(a+b+c)}$ and ${\displaystyle 2S=(A+B+C),}$

$${\displaystyle \begin{array}{rlrl}\sin \frac{1}{2}A& =\sqrt{\frac{\sin (s-b)\sin (s-c)}{\sin b\sin c}}& \sin \frac{1}{2}a& =\sqrt{\frac{-\cos S\cos (S-A)}{\sin B\sin C}}\\ \cos \frac{1}{2}A& =\sqrt{\frac{\sin s\sin (s-a)}{\sin b\sin c}}& \cos \frac{1}{2}a& =\sqrt{\frac{\cos (S-B)\cos (S-C)}{\sin B\sin C}}\\ \tan \frac{1}{2}A& =\sqrt{\frac{\sin (s-b)\sin (s-c)}{\sin s\sin (s-a)}}& \tan \frac{1}{2}a& =\sqrt{\frac{-\cos S\cos (S-A)}{\cos (S-B)\cos (S-C)}}\end{array}}$$

Another twelve identities follow by cyclic permutation.

The proof (Todhunter, Art.49) of the first formula starts from the identity ${\displaystyle 2{\sin }^2\frac{A}{2}=1-\cos A,}$ using the cosine rule to express A in terms of the sides and replacing the sum of two cosines by a product. (See sum-to-product identities.) The second formula starts from the identity ${\displaystyle 2{\cos }^2\frac{A}{2}=1+\cos A,}$ the third is a quotient and the remainder follow by applying the results to the polar triangle.

Delambre analogies

The Delambre analogies (also called Gauss analogies) were published independently by Delambre, Gauss, and Mollweide in 1807–1809.

$${\displaystyle \begin{array}{rlr}\frac{\sin \frac{1}{2}(A+B)}{\cos \frac{1}{2}C}=\frac{\cos \frac{1}{2}(a-b)}{\cos \frac{1}{2}c}& & \frac{\sin \frac{1}{2}(A-B)}{\cos \frac{1}{2}C}=\frac{\sin \frac{1}{2}(a-b)}{\sin \frac{1}{2}c}\\ \frac{\cos \frac{1}{2}(A+B)}{\sin \frac{1}{2}C}=\frac{\cos \frac{1}{2}(a+b)}{\cos \frac{1}{2}c}& & \frac{\cos \frac{1}{2}(A-B)}{\sin \frac{1}{2}C}=\frac{\sin \frac{1}{2}(a+b)}{\sin \frac{1}{2}c}\end{array}}$$

Another eight identities follow by cyclic permutation.

Proved by expanding the numerators and using the half angle formulae. (Todhunter, Art.54 and Delambre)

Napier's analogies

$${\displaystyle \begin{array}{rlr}\tan \frac{1}{2}(A+B)=\frac{\cos \frac{1}{2}(a-b)}{\cos \frac{1}{2}(a+b)}\cot \frac{1}{2}C& & \tan \frac{1}{2}(a+b)=\frac{\cos \frac{1}{2}(A-B)}{\cos \frac{1}{2}(A+B)}\tan \frac{1}{2}c\\ \tan \frac{1}{2}(A-B)=\frac{\sin \frac{1}{2}(a-b)}{\sin \frac{1}{2}(a+b)}\cot \frac{1}{2}C& & \tan \frac{1}{2}(a-b)=\frac{\sin \frac{1}{2}(A-B)}{\sin \frac{1}{2}(A+B)}\tan \frac{1}{2}c\end{array}}$$

Another eight identities follow by cyclic permutation.

These identities follow by division of the Delambre formulae. (Todhunter, Art.52)

Taking quotients of these yields the law of tangents, first stated by Persian mathematician Nasir al-Din al-Tusi (1201–1274),

$${\displaystyle \frac{\tan \frac{1}{2}(A-B)}{\tan \frac{1}{2}(A+B)}=\frac{\tan \frac{1}{2}(a-b)}{\tan \frac{1}{2}(a+b)}}$$

Napier's rules for right spherical triangles

When one of the angles, say C, of a spherical triangle is equal to π/2 the various identities given above are considerably simplified. There are ten identities relating three elements chosen from the set a, b, c, A, and B.

Napier provided an elegant mnemonic aid for the ten independent equations: the mnemonic is called Napier's circle or Napier's pentagon (when the circle in the above figure, right, is replaced by a pentagon).

First, write the six parts of the triangle (three vertex angles, three arc angles for the sides) in the order they occur around any circuit of the triangle: for the triangle shown above left, going clockwise starting with a gives aCbAcB. Next replace the parts that are not adjacent to C (that is A, c, and B) by their complements and then delete the angle C from the list. The remaining parts can then be drawn as five ordered, equal slices of a pentagram, or circle, as shown in the above figure (right). For any choice of three contiguous parts, one (the middle part) will be adjacent to two parts and opposite the other two parts. The ten Napier's Rules are given by

• sine of the middle part = the product of the tangents of the adjacent parts
• sine of the middle part = the product of the cosines of the opposite parts

The key for remembering which trigonometric function goes with which part is to look at the first vowel of the kind of part: middle parts take the sine, adjacent parts take the tangent, and opposite parts take the cosine. For an example, starting with the sector containing a we have:

$${\displaystyle \begin{array}{rl}\sin a& =\tan (\frac{\pi }{2}-B)\tan b\\ & =\cos (\frac{\pi }{2}-c)\cos (\frac{\pi }{2}-A)\\ & =\cot B\tan b\\ & =\sin c\sin A.\end{array}}$$

The full set of rules for the right spherical triangle is (Todhunter, Art.62)

$${\displaystyle \begin{array}{rlrlrlrl}& \text{(R1)}& \cos c& =\cos a\cos b,& & \text{(R6)}& \tan b& =\cos A\tan c,\\ & \text{(R2)}& \sin a& =\sin A\sin c,& & \text{(R7)}& \tan a& =\cos B\tan c,\\ & \text{(R3)}& \sin b& =\sin B\sin c,& & \text{(R8)}& \cos A& =\sin B\cos a,\\ & \text{(R4)}& \tan a& =\tan A\sin b,& & \text{(R9)}& \cos B& =\sin A\cos b,\\ & \text{(R5)}& \tan b& =\tan B\sin a,& & \text{(R10)}& \cos c& =\cot A\cot B.\end{array}}$$

Napier's rules for quadrantal triangles

A quadrantal spherical triangle together with Napier's circle for use in his mnemonics

A quadrantal spherical triangle is defined to be a spherical triangle in which one of the sides subtends an angle of π/2 radians at the centre of the sphere: on the unit sphere the side has length π/2. In the case that the side c has length π/2 on the unit sphere the equations governing the remaining sides and angles may be obtained by applying the rules for the right spherical triangle of the previous section to the polar triangle △A'B'C' with sides a', b', c' such that A' = π − a, a' = π − A etc. The results are:

$${\displaystyle \begin{array}{rlrlrlrl}& \text{(Q1)}& \cos C& =-\cos A\cos B,& & \text{(Q6)}& \tan B& =-\cos a\tan C,\\ & \text{(Q2)}& \sin A& =\sin a\sin C,& & \text{(Q7)}& \tan A& =-\cos b\tan C,\\ & \text{(Q3)}& \sin B& =\sin b\sin C,& & \text{(Q8)}& \cos a& =\sin b\cos A,\\ & \text{(Q4)}& \tan A& =\tan a\sin B,& & \text{(Q9)}& \cos b& =\sin a\cos B,\\ & \text{(Q5)}& \tan B& =\tan b\sin A,& & \text{(Q10)}& \cos C& =-\cot a\cot b.\end{array}}$$

Five-part rules

Substituting the second cosine rule into the first and simplifying gives:

$${\displaystyle \begin{array}{rl}\cos a& =(\cos a\cos c+\sin a\sin c\cos B)\cos c+\sin b\sin c\cos A\\ \cos a{\sin }^{2}c& =\sin a\cos c\sin c\cos B+\sin b\sin c\cos A\end{array}}$$

Cancelling the factor of sin c gives

$${\displaystyle \cos a\sin c=\sin a\cos c\cos B+\sin b\cos A}$$

Similar substitutions in the other cosine and supplementary cosine formulae give a large variety of 5-part rules. They are rarely used.

Cagnoli's Equation

Multiplying the first cosine rule by cos A gives

$${\displaystyle \cos a\cos A=\cos b\cos c\cos A+\sin b\sin c-\sin b\sin c{\sin }^{2}A.}$$

Similarly multiplying the first supplementary cosine rule by cos a yields

$${\displaystyle \cos a\cos A=-\cos B\cos C\cos a+\sin B\sin C-\sin B\sin C{\sin }^{2}a.}$$

Subtracting the two and noting that it follows from the sine rules that ${\displaystyle \sin b\sin c{\sin }^{2}A=\sin B\sin C{\sin }^{2}a}$ produces Cagnoli's equation

$${\displaystyle \sin b\sin c+\cos b\cos c\cos A=\sin B\sin C-\cos B\cos C\cos a}$$

which is a relation between the six parts of the spherical triangle.

Solution of triangles

Main article: Solution of triangles § Solving spherical triangles

Oblique triangles

The solution of triangles is the principal purpose of spherical trigonometry: given three, four or five elements of the triangle, determine the others. The case of five given elements is trivial, requiring only a single application of the sine rule. For four given elements there is one non-trivial case, which is discussed below. For three given elements there are six cases: three sides, two sides and an included or opposite angle, two angles and an included or opposite side, or three angles. (The last case has no analogue in planar trigonometry.) No single method solves all cases. The figure below shows the seven non-trivial cases: in each case the given sides are marked with a cross-bar and the given angles with an arc. (The given elements are also listed below the triangle). In the summary notation here such as ASA, A refers to a given angle and S refers to a given side, and the sequence of A's and S's in the notation refers to the corresponding sequence in the triangle.

• Case 1: three sides given (SSS). The cosine rule may be used to give the angles A, B, and C but, to avoid ambiguities, the half angle formulae are preferred.
• Case 2: two sides and an included angle given (SAS). The cosine rule gives a and then we are back to Case 1.
• Case 3: two sides and an opposite angle given (SSA). The sine rule gives C and then we have Case 7. There are either one or two solutions.
• Case 4: two angles and an included side given (ASA). The four-part cotangent formulae for sets (cBaC) and (BaCb) give c and b, then A follows from the sine rule.
• Case 5: two angles and an opposite side given (AAS). The sine rule gives b and then we have Case 7 (rotated). There are either one or two solutions.
• Case 6: three angles given (AAA). The supplemental cosine rule may be used to give the sides a, b, and c but, to avoid ambiguities, the half-side formulae are preferred.
• Case 7: two angles and two opposite sides given (SSAA). Use Napier's analogies for a and A; or, use Case 3 (SSA) or case 5 (AAS).

The solution methods listed here are not the only possible choices: many others are possible. In general it is better to choose methods that avoid taking an inverse sine because of the possible ambiguity between an angle and its supplement. The use of half-angle formulae is often advisable because half-angles will be less than π/2 and therefore free from ambiguity. There is a full discussion in Todhunter. The article Solution of triangles#Solving spherical triangles presents variants on these methods with a slightly different notation.

There is a full discussion of the solution of oblique triangles in Todhunter.^{Chap. VI} See also the discussion in Ross.

Solution by right-angled triangles

Another approach is to split the triangle into two right-angled triangles. For example, take the Case 3 example where b, c, and B are given. Construct the great circle from A that is normal to the side BC at the point D. Use Napier's rules to solve the triangle △ABD: use c and B to find the sides AD and BD and the angle ∠BAD. Then use Napier's rules to solve the triangle △ACD: that is use AD and b to find the side DC and the angles C and ∠DAC. The angle A and side a follow by addition.

Numerical considerations

Not all of the rules obtained are numerically robust in extreme examples, for example when an angle approaches zero or π. Problems and solutions may have to be examined carefully, particularly when writing code to solve an arbitrary triangle.

Area and spherical excess

See also: Solid angle and Geodesic polygon

Lexell's theorem: the triangles of constant area on a fixed base AB have their free vertex C along a small circle through the points antipodal to A and B.

Consider an N-sided spherical polygon and let A_n denote the n-th interior angle. The area of such a polygon is given by (Todhunter, Art.99)

$${\displaystyle \genfrac{}{}{0ex}{}{\text{Area of polygon}}{\text{(on the unit sphere)}}\equiv E_N=\left(\sum _{n=1}^N{A}_n\right)-(N-2)\pi .}$$

For the case of a spherical triangle with angles A, B, and C this reduces to Girard's theorem

$${\displaystyle \genfrac{}{}{0ex}{}{\text{Area of triangle}}{\text{(on the unit sphere)}}\equiv E=E_3=A+B+C-\pi ,}$$

where E is the amount by which the sum of the angles exceeds π radians, called the spherical excess of the triangle. This theorem is named after its author, Albert Girard. An earlier proof was derived, but not published, by the English mathematician Thomas Harriot. On a sphere of radius R both of the above area expressions are multiplied by R². The definition of the excess is independent of the radius of the sphere.

The converse result may be written as

$${\displaystyle A+B+C=\pi +\frac{4\pi \times \text{Area of triangle}}{\text{Area of the sphere}}.}$$

Since the area of a triangle cannot be negative the spherical excess is always positive. It is not necessarily small, because the sum of the angles may attain 5π (3π for proper angles). For example, an octant of a sphere is a spherical triangle with three right angles, so that the excess is π/2. In practical applications it is often small: for example the triangles of geodetic survey typically have a spherical excess much less than 1' of arc. (Rapp Clarke, Legendre's theorem on spherical triangles). On the Earth the excess of an equilateral triangle with sides 21.3 km (and area 393 km²) is approximately 1 arc second.

There are many formulae for the excess. For example, Todhunter, (Art.101—103) gives ten examples including that of L'Huilier:

$${\displaystyle \tan \frac{1}{4}E=\sqrt{\tan \frac{1}{2}s\tan \frac{1}{2}(s-a)\tan \frac{1}{2}(s-b)\tan \frac{1}{2}(s-c)}}$$

where ${\displaystyle s=\frac{1}{2}(a+b+c)}$

Because some triangles are badly characterized by their edges (e.g., if $a=b\approx \frac{1}{2}c$), it is often better to use the formula for the excess in terms of two edges and their included angle

$${\displaystyle \tan \frac{1}{2}E=\frac{\tan \frac{1}{2}a\tan \frac{1}{2}b\sin C}{1+\tan \frac{1}{2}a\tan \frac{1}{2}b\cos C}.}$$

When triangle △ABC is a right triangle with right angle at C, then cos C = 0 and sin C = 1, so this reduces to

$${\displaystyle \tan \frac{1}{2}E=\tan \frac{1}{2}a\tan \frac{1}{2}b.}$$

Angle deficit is defined similarly for hyperbolic geometry.

From latitude and longitude

The spherical excess of a spherical quadrangle bounded by the equator, the two meridians of longitudes ${\displaystyle {\lambda }_1}$ and ${\displaystyle {\lambda }_2,}$ and the great-circle arc between two points with longitude and latitude ${\displaystyle ({\lambda }_1,{\phi }_1)}$ and ${\displaystyle ({\lambda }_2,{\phi }_2)}$ is

$${\displaystyle \tan \frac{1}{2}{E}_4=\frac{\sin \frac{1}{2}({\phi }_2+{\phi }_1)}{\cos \frac{1}{2}({\phi }_2-{\phi }_1)}\tan \frac{1}{2}({\lambda }_2-{\lambda }_1).}$$

This result is obtained from one of Napier's analogies. In the limit where ${\displaystyle {\phi }_1,{\phi }_2,{\lambda }_2-{\lambda }_1}$ are all small, this reduces to the familiar trapezoidal area, $E_4\approx \frac{1}{2}({\phi }_2+{\phi }_1)({\lambda }_2-{\lambda }_1)$.

The area of a polygon can be calculated from individual quadrangles of the above type, from (analogously) individual triangle bounded by a segment of the polygon and two meridians, by a line integral with Green's theorem, or via an equal-area projection as commonly done in GIS. The other algorithms can still be used with the side lengths calculated using a great-circle distance formula.

Surface area

From Wikipedia, the free encyclopedia

A sphere of radius r has surface area 4πr².

The surface area (symbol A) of a solid object is a measure of the total area that the surface of the object occupies. The mathematical definition of surface area in the presence of curved surfaces is considerably more involved than the definition of arc length of one-dimensional curves, or of the surface area for polyhedra (i.e., objects with flat polygonal faces), for which the surface area is the sum of the areas of its faces. Smooth surfaces, such as a sphere, are assigned surface area using their representation as parametric surfaces. This definition of surface area is based on methods of infinitesimal calculus and involves partial derivatives and double integration.

A general definition of surface area was sought by Henri Lebesgue and Hermann Minkowski at the turn of the twentieth century. Their work led to the development of geometric measure theory, which studies various notions of surface area for irregular objects of any dimension. An important example is the Minkowski content of a surface.

Definition

While the areas of many simple surfaces have been known since antiquity, a rigorous mathematical definition of area requires a great deal of care. This should provide a function

${\displaystyle S\mapsto A(S)}$

which assigns a positive real number to a certain class of surfaces that satisfies several natural requirements. The most fundamental property of the surface area is its additivity: the area of the whole is the sum of the areas of the parts. More rigorously, if a surface S is a union of finitely many pieces S₁, …, S_r which do not overlap except at their boundaries, then

${\displaystyle A(S)=A(S_1)+\cdots +A(S_r).}$

Surface areas of flat polygonal shapes must agree with their geometrically defined area. Since surface area is a geometric notion, areas of congruent surfaces must be the same and the area must depend only on the shape of the surface, but not on its position and orientation in space. This means that surface area is invariant under the group of Euclidean motions. These properties uniquely characterize surface area for a wide class of geometric surfaces called piecewise smooth. Such surfaces consist of finitely many pieces that can be represented in the parametric form

${\displaystyle S_D:\overrightarrow{r}=\overrightarrow{r}(u,v),(u,v)\in D}$

with a continuously differentiable function ${\displaystyle \overrightarrow{r}.}$ The area of an individual piece is defined by the formula

${\displaystyle A(S_D)={\iint }_D\left|{\overrightarrow{r}}_u\times {\overrightarrow{r}}_v\right|dudv.}$

Thus the area of S_D is obtained by integrating the length of the normal vector ${\displaystyle {\overrightarrow{r}}_u\times {\overrightarrow{r}}_v}$ to the surface over the appropriate region D in the parametric uv plane. The area of the whole surface is then obtained by adding together the areas of the pieces, using additivity of surface area. The main formula can be specialized to different classes of surfaces, giving, in particular, formulas for areas of graphs z = f(x,y) and surfaces of revolution.

Schwarz lantern with ${\displaystyle M}$ axial slices and ${\displaystyle N}$ radial vertices. The limit of the area as ${\displaystyle M}$ and ${\displaystyle N}$ tend to infinity doesn't converge. In particular it doesn't converge to the area of the cylinder.

One of the subtleties of surface area, as compared to arc length of curves, is that surface area cannot be defined simply as the limit of areas of polyhedral shapes approximating a given smooth surface. It was demonstrated by Hermann Schwarz that already for the cylinder, different choices of approximating flat surfaces can lead to different limiting values of the area; this example is known as the Schwarz lantern.

Various approaches to a general definition of surface area were developed in the late nineteenth and the early twentieth century by Henri Lebesgue and Hermann Minkowski. While for piecewise smooth surfaces there is a unique natural notion of surface area, if a surface is very irregular, or rough, then it may not be possible to assign an area to it at all. A typical example is given by a surface with spikes spread throughout in a dense fashion. Many surfaces of this type occur in the study of fractals. Extensions of the notion of area which partially fulfill its function and may be defined even for very badly irregular surfaces are studied in geometric measure theory. A specific example of such an extension is the Minkowski content of the surface.

Common formulas

See also: List of formulas in elementary geometry

 

Surface areas of common solids

Shape	Equation	Variables
Cube	${\displaystyle 6a^2}$	a = side length
Cuboid	${\displaystyle 2\left(lb+lh+bh\right)}$	l = length, b = breadth, h = height
Triangular prism	${\displaystyle bh+l\left(p+q+r\right)}$	b = base length of triangle, h = height of triangle, l = distance between triangular bases, p, q, r = sides of triangle
All prisms	${\displaystyle 2B+Ph}$	B = the area of one base, P = the perimeter of one base, h = height
Sphere	${\displaystyle 4\pi r^2=\pi d^2}$	r = radius of sphere, d = diameter
Hemisphere	${\displaystyle 3\pi r^2}$	r = radius of the hemisphere
Hemispherical shell	${\displaystyle \pi \left(3R^2+r^2\right)}$	R = external radius of hemisphere, r = internal radius of hemisphere
Spherical lune	${\displaystyle 2r^2\theta }$	r = radius of sphere, θ = dihedral angle
Torus	${\displaystyle \left(2\pi r\right)\left(2\pi R\right)=4{\pi }^{2}Rr}$	r = minor radius (radius of the tube), R = major radius (distance from center of tube to center of torus)
Closed cylinder	${\displaystyle 2\pi r^2+2\pi rh=2\pi r\left(r+h\right)}$	r = radius of the circular base, h = height of the cylinder
Cylindrical annulus	${\displaystyle 2\pi Rh+2\pi rh+2(\pi R^2-\pi r^2)=2\pi (R+r)(R-r+h)}$	R = External radius r = Internal radius, h = height
Capsule	${\displaystyle 2\pi r(2r+h)}$	r = radius of the hemispheres and cylinder, h = height of the cylinder
Curved surface area of a cone	${\displaystyle \pi r\sqrt{r^2+h^2}=\pi rs}$	${\displaystyle s=\sqrt{r^2+h^2}}$ s = slant height of the cone, r = radius of the circular base, h = height of the cone
Full surface area of a cone	${\displaystyle \pi r\left(r+\sqrt{r^2+h^2}\right)=\pi r\left(r+s\right)}$	s = slant height of the cone, r = radius of the circular base, h = height of the cone
Regular Pyramid	${\displaystyle B+\frac{Ps}{2}}$	B = area of base, P = perimeter of base, s = slant height
Square pyramid	${\displaystyle b^2+2bs=b^2+2b\sqrt{{\left(\frac{b}{2}\right)}^2+h^2}}$	b = base length, s = slant height, h = vertical height
Rectangular pyramid	${\displaystyle lb+l\sqrt{{\left(\frac{b}{2}\right)}^2+h^2}+b\sqrt{{\left(\frac{l}{2}\right)}^2+h^2}}$	l = length, b = breadth, h = height
Tetrahedron	${\displaystyle \sqrt{3}{a}^2}$	a = side length
Surface of revolution	${\displaystyle 2\pi {\int }_a^{b}f(x)\sqrt{1+(f^{\prime }(x){)}^2}dx}$
Parametric surface	${\displaystyle {\iint }_D\left|{\overrightarrow{r}}_u\times {\overrightarrow{r}}_v\right|dA}$	${\displaystyle \overrightarrow{r}}$ = parametric vector equation of surface, ${\displaystyle {\overrightarrow{r}}_u}$ = partial derivative of ${\displaystyle \overrightarrow{r}}$ with respect to ${\displaystyle u}$, ${\displaystyle {\overrightarrow{r}}_v}$ = partial derivative of ${\displaystyle \overrightarrow{r}}$ with respect to ${\displaystyle v}$, ${\displaystyle D}$ = shadow region

Ratio of surface areas of a sphere and cylinder of the same radius and height

A cone, sphere and cylinder of radius r and height h.

The below given formulas can be used to show that the surface area of a sphere and cylinder of the same radius and height are in the ratio 2 : 3, as follows.

Let the radius be r and the height be h (which is 2r for the sphere).

$${\displaystyle \begin{array}{rlll}\text{Sphere surface area}& =4\pi r^2& & =(2\pi r^2)\times 2\\ \text{Cylinder surface area}& =2\pi r(h+r)& =2\pi r(2r+r)& =(2\pi r^2)\times 3\end{array}}$$

The discovery of this ratio is credited to Archimedes.

In chemistry

Surface area of particles of different sizes.

See also: Accessible surface area

Surface area is important in chemical kinetics. Increasing the surface area of a substance generally increases the rate of a chemical reaction. For example, iron in a fine powder will combust, while in solid blocks it is stable enough to use in structures. For different applications a minimal or maximal surface area may be desired.

In biology

The inner membrane of the mitochondrion has a large surface area due to infoldings, allowing higher rates of cellular respiration (electron micrograph).

The surface area of an organism is important in several considerations, such as regulation of body temperature and digestion. Animals use their teeth to grind food down into smaller particles, increasing the surface area available for digestion. The epithelial tissue lining the digestive tract contains microvilli, greatly increasing the area available for absorption. Elephants have large ears, allowing them to regulate their own body temperature. In other instances, animals will need to minimize surface area; for example, people will fold their arms over their chest when cold to minimize heat loss.

The surface area to volume ratio (SA:V) of a cell imposes upper limits on size, as the volume increases much faster than does the surface area, thus limiting the rate at which substances diffuse from the interior across the cell membrane to interstitial spaces or to other cells. Indeed, representing a cell as an idealized sphere of radius r, the volume and surface area are, respectively, V = (4/3)πr³ and SA = 4πr². The resulting surface area to volume ratio is therefore 3/r. Thus, if a cell has a radius of 1 μm, the SA:V ratio is 3; whereas if the radius of the cell is instead 10 μm, then the SA:V ratio becomes 0.3. With a cell radius of 100, SA:V ratio is 0.03. Thus, the surface area falls off steeply with increasing volume.