Spherical Trigonometry

1 Definitions: Geometrical Properties of the Sphere and

Spherical Triangles

1.1 Great and small circles (see figure 1)

• A great circle on the surface of a sphere is one whose plane passes through the center ofthe sphere. For example, all meridians and the Equator. A great circle divides the sphereinto two equal parts.

• A small circle is one whose plane does not passes through the center of the sphere. Aparallel of latitude, such as LL′ is a small circle.

• The pole of a circle is that point on the surface of the sphere which is equidistant fromall points on the circle.

• The angular distance, or arc, PL is called the spherical radius of a small circle LL′, andP is called the pole of that small circle. Otherwise, the radius of a small circle is usuallytaken to mean OL.

• P is also the pole of the great circle QQ′, and clearly, the spherical radius of a great circleis a quadrant, that is 90◦. It is also clear that all great circle have two poles.

• Lune is a portion of the surface of a sphere enclosed by two great circles. For instance,PLQSAP is a lune, and ∠LPA is the angle of the lune.

L L′

O

P

S

Q Q′

A

figure 1

1.2 Angle between planes of two great circles (see figure 2)

• B is the pole of the great circle DC and A is the pole of the great circle EC.

• BO is normal to plane ODC and AO is normal to the plane OEC.

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• DO and EO are both normal to the line of intersection of the planes, CO.

• Hence, the angle between the planes of two great circles can be defined as the arc of agreat circle between the poles of the two great circle whose planes we are considering, i.e.,∠EOD. Also, ∠EOD = ∠BOA.

B

A

OD

E

C

T

T ′figure 2

1.3 Spherical angle between two great circle (see figure 2)

• CT is the tangent to the great circle CD at C and CT ′ is the tangent to the great circleCE at C.

• The spherical angle ∠DCE is defined as the angle between the tangents to the great circlesdrawn at the point of intersection, i.e., ∠TCT ′ is the spherical angle between the greatcircles at C, and clearly

∠TCT ′ = angle between the planes = ∠EOD.

1.4 Spherical triangle (see figure 3a)

• Spherical triangle is defined as a triangle formed on the surface of a sphere by the intersec-tion of the arcs of three great circles, for example, PAB. Note that the sides are formedonly by arcs of great circles, never by small circles.

• Similarly, a spherical quadrilateral is formed by the arcs of four great circles.

• The arcs enclosing the spherical triangle are spoken of as its sides, and the angles in whichthese arcs intersect are the angles of the triangle. The three sides and three angles arecollectively termed the parts of a spherical triangle, a term which is useful occasionally,since in reality all six are angles.

• The length of side of a spherical triangle is defined as the angle subtended by that side atthe center of the sphere. For example, length of side PB is ∠POB; side PA = ∠POA,and so on. Thus, the sides like the angles, are expressed in degrees, minutes and secondsof arc.

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• If the actual length of the side is required this may easily be founded knowing the radiusof the sphere. For example,

length of AB = R× θ,

where R is the radius of the sphere and θ = ∠AOB expressed in radians.

O

P

AB

O

P

AB

Q

figure 3a figure 3b

1.5 Size of a spherical triangle (see figure 3b)

• The arc joining two points is , by convention, always taken to mean the lesser segment ofthe great circle passing through the two points, i.e., in the triangle PAB by the arc PA,we mean the obvious one, not PQA.

• Each side of a spherical triangle must be less than 180◦.

• Each angle of a spherical triangle must be less than 180◦.

• The sum of the three sides must lie between 0◦ to 360◦.

• The sum of the three angles must lie between 180◦ to 540◦.

• The area of any spherical triangle must be less than 2πR2, where R is the radius of thesphere.

1.6 Some useful geometrical properties

• Any two sides of a spherical triangle are together greater than the third.

• In any spherical triangle, the greater angle is opposite to the greater side, and the converseis also true, i.e., the greater side is opposite to the greater angle.

• The angles at the base of an isosceles triangle are equal (see figure 4.)

3

=

=

figure 4

1.7 Spherical excess

It is easy to see that the sum of the three angles of a spherical triangle is not always 180◦. Itlies between 180◦ and 540◦. The extra value is called spherical excess.

For example, an octant is an example of a spherical triangle with all its angles being 90◦. Thus,the spherical excess is 90◦ (see figure 5.)

figure 5

2 Application of Spherical Trigonometry

2.1 Terrestrial Sphere

If we assume the earth to be a perfect sphere, the following section gives a summary of theprincipal terms used, without, however, attempting to define them in any way precisely.

A

N

P

D

B T

M

O

V

figure 6

4

Referring to figure 6, in which A and B are two places on the surface of the earth, P is the poleof the earth, O is the center of the earth. Then we can define the following terms:

• Arc AN (or ∠AON) = latitude of A.

• Arc BM (or ∠BOM) = latitude of B.

• PA = co-latitude of A (i.e., complement of the latitude of A = 90◦ − ∠AON .)

• PB = co-latitude of B

• Length of arc AB (i.e., ∠AOB) = the great circle distance from A to B.

• PN , PM = meridians of A and B, respectively.

• Arc NM (or ∠NPM) = difference of longitude between A and B.

Also, for a vessel on track from A towards B,

∠PAB = initial course (or track angle);

∠PBT = final course (or track angle).

V is the vertex of the great circle, i.e., the point at which the great circle is the nearest to thepole, so that PV is at right angles to the great circle of AB.

Also AD is the distance along a parallel of latitude between the two meridians and when ex-pressed in nautical miles, is called the departure.

3 Basic Formulas

3.1 The cosine formula

The formula states that in any spherical triangle (see figure 7,)

cos a = cos b cos c+ sin b sin c cosA.

O

A

B

C

D

E

a

bc

ab

c

Figure 7

5

Proof. Let the triangle ABC be as shown in figure 6, O is the center of the sphere. Side a, b,c, are measured by the angles at the center of the sphere, as indicated.

AE and AD are perpendicular to OA, and ∠EAD = ∠A.

In the plane triangle AED (by the cosine formula,)

DE2 = AE2 + AD2− 2AE · AD cosA, (1)

and in triangle OED,DE2 = OE2 +OD2

− 2OD ·OD cos a. (2)

Also,OE2 = AE2 +OA2 and OD2 = AD2 +OA2,

so that (2) becomes

DE2 = 2OA2 + AE2 + AD2− 2OD ·OD cos a (3)

and equating (1) and (3).

=⇒2OA2− 2OE ·OD cos a = −2AE · AD cosA

=⇒OE ·OD cos a = OA2 + AE · AD cosA

=⇒ cos a =OA

OE·OA

OD+

AE

OE·AD

ODcosA

=⇒ cos a = cos b cos c+ sin b sin c cosA

It follows without further proof that

cos b = cos a cos c+ sin a sin c cosB;

cos c = cos a cos b+ sin a sin b cosC.

Note:

• The above is the usual proof of the cosine formula. It does assume that the two sidesincluding the given angle, AB and AC, are both less than 90◦. Generalization to largerthan 90◦ is not difficult.

• In the form stated above, it may be used, given two sides and the included angle, to findthe third side.

• By transposing the formula, we get

cosA =cos a− cos b cos c

sin b sin c

a form in which it may be used, given three sides, to find any angle.

Example 3.1 In △PXZ, ∠P = 50◦, z = 70◦45′, x = 62◦10′, find p and ∠Z.

cos p =cos x cos z + sin x sin z cosP

=cos 62◦10′ cos 70◦45′ + sin 62◦10′ sin 70◦45′ cos 50◦ ≈ 0.6906

=⇒ p =46◦19′

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cosZ =cos z − cos p cos x

sin p sin x

=cos 70◦45′ − cos 46◦19′ cos 62◦10′

sin 46◦19′ sin 62◦10′≈ 0.01128

=⇒ Z =89◦21′

3.2 The sine formula

This states that in any spherical triangle (see figure 8,)

sin a

sinA=

sin b

sinB=

sin c

sinC.

O

A

B

C

DE

F

P

b

cc

b

C

figure 8

Proof. Let ABC be a spherical triangle, and O be the center of this sphere. Let P be anypoint on OA. From P , drop perpendicular PD to the plane OBC. Drop perpendiculars DF

and DE to OC and OB, respectively. Join PF , PE and OD.

Then ∠PDF , ∠PDE and ∠PDO are all right angles, and by construction ∠DFO, ∠DEO arealso right angles.

To show that ∠PFO is also a right angle, we have

PF 2 =PD2 +DF 2

=(

PO2−OD2

)

+(

OD2−OF 2

)

,

i.e., PF 2 = PO2−OF 2, so △PFO is right angled at F . Similarly, △PEO is right angled at E.

We now have, PF = PO sin b and also

PD = PF sinC = PO sin b sinC. (4)

Similarly, PE = PO sin c and

PD = PE sinB = PO sin c sinB. (5)

Equating (4) and (5)

sin b sinC = sin c sinB =⇒sin b

sinB=

sin c

sinC.

Similarly, by dropping a perpendicular from a point in OB onto plane OAC, it may be shownthat

sin a

sinA=

sin c

sinC.

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Hence,sin a

sinA=

sin b

sinB=

sin c

sinC.

Note:

• The sine formula gives the relation between two angles and the two sides opposite to them.Given any three of these parts, the fourth may be found.

• One big disadvantage of the sine rule is due to the fact that sinA = sin (180◦ − A), i.e.,both A and (180◦ − A) are possible candidates to the solution of our equation. Usuallywe overcome this ambiguity by using:

i. Physical setting of the problem.

ii. Greater side faces greater angle and hence (A−B) and (a− b) must be of the samesign.

Example 3.2 In △PAB, P is the pole and A, B are two place in the northern hemisphere.Given ∠A = 68◦, AB = p = 60◦30′, ∠P = 80◦16′, find the latitude of B.

sin a

sinA=

sin p

sinP=⇒ sin a =

sinA sin p

sinP≈ 0.8188

=⇒a = 54◦58′

Hence, the latitude of B is 35◦02′N.

Note: since∠P > ∠A =⇒ p > a,

then 54◦58′ is the only choice, but not 125◦02′.

Example 3.3 Consider a practical example, a triangle PAB in which b = 26◦21′, B = 52◦22′,A = 104◦44′.

To find a, we have

sin a

sinA=

sin b

sinB=⇒ sin a =

sin b sinA

sinB=⇒a = 32◦49′ or a = 147◦10′

Since (A− B) is positive, and so (a− b) must also be positive. However, both values for a satisfythis requirement; thus both are solutions of the data as given, and from the practical point ofview we are not much further ahead.

Example 3.4 Two places A and B on the earth have the follow latitudes and longitudes:A (40◦N, 18◦E) and B (0◦N, 58◦E). Find the angle of departure from A to B of the great circleroute.

8

N

A

Bequator

primemeridian

figure 9

By the given latitudes and longitudes, AN = 50◦, BN = 90◦ and ∠ANB = 58◦ − 18◦ = 40◦.Using the cosine formula,

cosAB =cosAN cosBN + sinAN sinBN cos∠ANB

=cos 50◦ cos 90◦ + sin 50◦ sin 90◦ cos 40◦ ≈ 0.5868

=⇒ AB =54◦04′.

Then by sine formula,

sin∠NAB

sinBN=

sin∠ANB

sinAB=⇒ sin∠NAB =

sin∠ANB sinBN

sinAB=

sin 40◦ sin 90◦

sin 54◦4′≈ 0.7939

=⇒∠NAB = 52◦33′ or 127◦27′.

From the physical setting, we know that ∠NAB = 127◦27′.

3.3 The polar triangle and supplemental theorem

3.3.1 The polar triangle

A

B

C

A′

B′

C ′

ab

c

figure 10a

By the pole of a great circle we mean the point that is 90◦ distant from every point on the greatcircle.

Let ABC be a spherical triangle in which sides a, b, c are parts of the great circle suggested bytheir respective curvatures.

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Let A′ be the pole of the great circle which includes a.

Let B′ be the pole of the great circle which includes b.

Let C ′ be the pole of the great circle which includes c.

Then A′B′C ′ is called the polar triangle of ABC, and ABC is called the primitive triangle.

Note that since every great circle has two poles, a series of such triangles might be formed byjoining the poles in different ways. By convention, the poles chosen must lie on the same sideof the arc as the opposite angle. For instance, A′ lies in the same direction from arc a as doesA, and so do B′ and B, and C ′ and C, in relation to their respective arcs.

This is the only triangle which is referred to as the polar triangle.

If A′B′C ′ is the polar triangle of ABC, then it can be shown that ABC is the polar triangle ofA′B′C ′.

Let A′ lie upon the same side of BC (arc a) as A.

Then if A′ is the pole of arc a, we can show that A is the pole of arc a′.

Proof.

A

B

C

A′

B′

C ′

ab

c

a′

b′

c′

figure 10b

Join AC ′, AB′. Since B′ is the pole of AC, then AB′ is a quadrant. Since C ′ is the pole of AB,then AC ′ is a quadrant.

Hence, since both AC ′ and AB′ are quadrants, it follows that A is the pole of arc a′. Similarly,B is the pole of b′ and C is the pole of c′.

3.3.2 The supplemental theorem

This is a very important theorem relating to polar triangles. It states that:

The angles in the polar triangle are supplements of the corresponding sides in the primitivetriangle, and the sides in the polar triangle are supplements of the corresponding angles in theprimitive triangle.

Note: “corresponding” in the sense that we call b the side in the primitive corresponding to B′

in the polar, and so on.

Hence, the supplemental theorem states that:

B′ = 180◦ − b; a = 180◦ − A′; or in radians, c′ = π − C, and so on.

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Proof.

A

B

C

A′

B′

C ′

D

E

figure 11

Let side A′C ′ intersect BA and BC in D and E, respectively.

Since B is the pole of A′C ′, i.e., the pole of DE, both BD and BE are quadrants and so arcDE = ∠B.

Since A′ is the pole of BC, then A′E is a quadrant.

Since C ′ is the pole of AB, then C ′D is a quadrant.

So,

C ′D + A′E = 180◦

=⇒A′C ′ +DE = 180◦

=⇒A′C ′ + ∠B = 180◦.

Hence, ∠B and side A′C ′ (i.e. b′) are supplements.

Similarly, it may be shown that A and a′, C and c′ are supplements.

Example 3.5 In a triangle PZX, given x = 55◦14′, P = 54◦01′, Z = 121◦25′, solve for X.

As it stands, two angles and included side. So, in the polar triangle,

X ′ = 180◦ − x = 124◦46′, p′ = 180◦ − P = 125◦59′, z′ = 180◦ − Z = 58◦35′.

cosx′ = cos p′ cos z′ + sin p′ sin z′ cosX ′ =⇒ cos x′≈ −0.700047

=⇒x′ = 134◦26′

=⇒X = 180◦ − x′ = 45◦34′

3.4 Supplemental formulas

By applying to the polar triangle A′B′C ′ any formula connecting sides and angles, a supplementalformula may be obtained for triangle ABC involving the sides and angles opposite to the anglesand sides which appear in the original formula. For example, consider

cos a = cos b cos c+ sin b sin c cosA.

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If we apply it to the supplemental triangle, we have

cos a′ = cos b′ cos c′ + sin b′ sin c′ cosA′.

Since a′ = 180◦ − A, etc., this gives

cos (180◦ − A) = cos (180◦ − B) cos (180◦ − C) + sin (180◦ −B) sin (180◦ − C) cos (180◦ − a)

=⇒ − cosA = (− cosB) (− cosC) + sinB sinC (− cos a)

=⇒ cosA = − cosB cosC + sinB sinC cos a.

Example 3.6 Using the supplemental formula to solve for Example 3.5 again,

cosX = − cosP cosZ + sinP sinZ cos x

= − cos 54◦01′ cos 121◦25′ + sin 54◦01′ sin 121◦25′ cos 55◦14′

=⇒ x = 45◦34′.

Important:

Giving any three quantities of a spherical triangle (which has six quantities, the three sides andthree angles,) we can solve the triangle.

Summary of formulas used

Condition for defining the spherical triangle Suitable formulaSSS, SAS Cosine formulaAAS Sine formulaASA, AAA Supplemental formulas for polar triangle

Exercises

1. Solve the following spherical triangles:

(a) PAB, given a = 57◦00′, B = 94◦01′, P = 71◦51.5′, find b and p.

Ans. b = 83◦00′, p = 71◦00′

(b) PZX, given P = 63◦47.5′, Z = 83◦56.5′, X = 93◦34′, find p and z.

Ans. p = 64◦00′, z = 85◦00′

(c) PZY , given Z = 70◦27′, P = 114◦54′, Y = 109◦42′, find p and z.

Ans. p = 127◦00′, z = 56◦05′

(d) PZX, given p = 87◦10′, z = 62◦37′, x = 100◦10′, find P and Z.

Ans. P = 81◦24′, Z = 61◦32′

(e) PZX, given P = 88◦24.5′, z = 98◦10′, x = 100◦09′, find p and Z.

Ans. p = 87◦01′, Z = 97◦46′

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(f) PAB, given A = 33◦14′, a = 80◦05′, b = 70◦12′, find B.

Ans. B = 31◦34′

(g) PZX, given P = 51◦30′, Z = 113◦30′, PZ = 60◦20′, find PX and ZX.

Ans. PX = 92◦07′, ZX = 58◦31′

(h) ABC, given a = 49◦08′, C = 71◦20′, b = 58◦23′, find A and B.

Ans. A = 59◦01′, B = 74◦52′

(i) PZX, given P = 85◦30′, PZ = 49◦34′, PX = 99◦58′, find Z.

Ans. Z = 100◦30′

(j) ABC, given A = 88◦36′, B = 121◦36′, C = 69◦35′, find a and b.

Ans. a = 101◦27′, b = 123◦23′

4 Right-angled and Quadrantal Triangle and Napier’s

Rules

Definition 4.1 If a spherical triangle has one of its angles equal to 90◦, it is called a right-

angled triangle and if it has one of its sides equal to 90◦ that is to say, a quadrant, it is saidto be a quadrantal triangle.

In such circumstances, the solution of the triangle becomes a much simpler matter.

Consider the spherical triangle ABC, in which ∠C = 90◦, and let us select formula which include∠C.

A

BC a

b

c

figure 12

For instance, by the cosine formula,

cos c = cos a cos b+ sin a sin b cosC = cos a cos b

since cos 90◦ = 0.

4.1 Napier’s rules

Napier established that formulas applicable to right-angled spherical triangle can be obtainedfrom an arrangement that can be easily memorized.

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AB

C

a b

c

c

Bc

acCc

b

figure 13

As shown in figure 13, five angles are set around a circle with the two sides containing the rightangle and the complements of the other three elements. The complement of an element is thedifference between 90◦ and the element, i.e., ac = 90◦ − a. If we select any one of these fivecircular parts, as they are called the following relations hold:

sine of middle part = product of tangents of adjacent parts

= product of cosines of opposite parts

Note that the sine, cosine and tangent of the complement of the angle or the side θ are

sin θc = sin (90◦ − θ) = cos θ;

cos θc = cos (90◦ − θ) = sin θ;

tan θc =sin (90◦ − θ)

cos (90◦ − θ)=

cos θ

sin θ= cot θ.

Example 4.1 Consider the right-angled triangle in figure 13.

For the adjacent parts, angle B, side a and side c,

sinBc = tan ac tan c

=⇒ sin (90◦ −B) = tan (90◦ − a) tan c

=⇒ cosB = cot a tan c.

For the opposite parts, angle B, angle C and side b,

sinBc = cosCc cos b

=⇒ cosB = sinC cos b.

Example 4.2 If ∠C = 90◦, a = 116◦ and b = 32◦, find ∠A.

BA

C

b = 32◦ a = 116◦

c

a

Bc

ccAc

b

figure 12

14

Since A, b and a are the adjacent parts, then

sin b = tanAc tan a

=⇒ sin b = cotA tan a

=⇒ tanA =tan a

sin b=

tan 116◦

sin 32◦

=⇒A = 104◦30′.

Example 4.3 A ship leaves a place (lat. 35◦40′N, long. 141◦E) and follows a great circle track.If the initial course is 60◦30′, find the latitude of the vertex, i.e., the point with the shortestdistance from the pole.

The path that gives the shortest distance between the pole and the course of the ship travellingis the great circle passing through the pole and perpendicular to the travelling course of the ship.

A

P

V

av

P c

a

p

Ac

vc

figure 13

So, ∠V = 90◦, A = 60◦30′ and v = 90◦ − 35◦40′ = 54◦20′. By Napier’s rule,

sin a = cosAc cos vc = sinA sin v = sin 60◦30′ sin 54◦20′

=⇒a = 45◦00′ or 135◦00′.

Since ∠A is less than 90◦, so a must also be less than 90◦, based on the physical setting. Hence,the latitude of V is

90◦ − 45◦00′ = 45◦00′N.

Example 4.4 In △PXY , p = 53◦20′, X = 92◦05′, Y = 90◦. Find the other parts.

P c

yc

Xc

p

x

figure 14

15

For y,

sinXc = tan yc tan p

=⇒ tan y =tanP

cosX=

tan 53◦20′

cos 92◦05′=⇒ y = 91◦33′.

For P ,

sinP c = cos p cosXc

=⇒ cosP = cos p sinX = cos 53◦20′ sin 92◦05′ =⇒ P = 53◦22′.

For x,

sin x = tan p tanP c

=⇒ sin x = tan p cotP = tan 53◦20′ cot 53◦22′ =⇒ x = 87◦11′ or 92◦49′.

Since ∠X > ∠Y , then x > y, i.e., x = 92◦49′.

4.2 Quadrantal triangle

Napier’s Rules apply to quadrantal triangle with one important modification, which must benoted:

Theorem 4.1 In a quadrantal triangle, if both adjancents or both opposites are both

sides or both angles, put in a minus sign (i.e. put a minus sign in front of the product.)

In △ABC, let c (AB) be the quadrant (90◦.) Then A and B are next to the quadrant, theother three parts are written down as complements. For example, given b and C, find A. (b isthe middle part, the other two are adjacent, and are both angles, so put in a minus sign.)

A

B

Ca

bc = 90◦

bc

Cc

acB

A

figure 15

By Napier’s Rules,

sin bc = − tanA tanCc

=⇒ cos b = − tanA cotC.

This means for instance, that if b and C are both less than 90◦, then A is greater than 90◦ sincetanA is negative.

Or again, for B

sinB = cos bc cosCc = sin b sinC (no minus sign this time.)

16

Example 4.5 In the quadrantal triangle ABC in which a = 90◦, b = 78◦14′ and c = 49◦08′,find A and B.

A

B Ca = 90◦

b = 78◦14′c = 49◦08′bc

C

B

cc

Ac

figure 16

Since A is the middle part with b and c are the adjacent parts, then

sinAc = − tan bc tan cc

=⇒ cosA = − cot b cot c = − cot 78◦14′ cot 49◦08′ =⇒ A = 100◦23′.

Since b is the middle part with B and c are the opposite parts, then

sin bc = cosB cos cc

=⇒ cosB =cos b

sin c=

cos 78◦14′

sin 49◦08′=⇒ B = 74◦21′.

Exercises

1. Solve the spherical triangle ABC in which A = 90◦, a = 75◦49′ and b = 46◦12′.

Ans. B = 48◦07′, c = 69◦16′, C = 74◦43′

2. Solve the spherical triangle ABC in which C = 90◦, a = 109◦16′ and B = 38◦45′.

Ans. b = 37◦09′, c = 105◦15′, A = 101◦55′

3. The initial course for a great circle track from New York (lat. 42◦42′N, long.74◦01′W) is30◦10′. Find the position on the track which is neartest to the North pole. Find also thedistance along the track.

Ans. 68◦20′N, 5◦32′W, 2588.082 nautical miles

4. Find the shortest distance between two position in latitude 49◦50′N if their difference oflongitude is 50◦.

Ans. 1898.27 nautical miles

5. Solve the spherical triangle ABC in which a = 90◦, B = 45◦ and c = 72◦.

Ans. C = 65◦20′, b = 47◦44′, A = 107◦10′

6. Solve the spherical triangle XY Z in which X = 73◦01′, XZ = 47◦47′ and Y Z = 90◦.

Ans. Y = 45◦06′, Z = 114◦27′, XY = 107◦52′

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7. A vessel in lat. 40◦S, long. 30◦E, sails along a great circle track. If the initial course is128◦, and the distance steamed is 5400 nautical miles, find the position reached.

Ans. 28◦08′S, 146◦40′E

8. Find the shortest distance between city A in lat. 48◦24′N and long. 124◦44′W, and cityB in lat. 34◦50′N and long. 139◦50′W.

Ans. 1054 nautical miles

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