SPHERICAL TRIGONOMETRY
FOR
COLLEGES AND SECONDARY SCHOOLS
BY
DANIEL A. MUjRRAY, B.A., PH.D.
INSTRUCTOR IN MATHEMATICS IN CORNELL UNIVERSITY
FORMERLY SCHOLAR AND FELLOW AT JOHNS HOPKINS UNIVERSITY
LONGMANS, GRjEEN, AND CO.
91 AND 93 FIFTH AVENUE, NEW YORK
LONDON AND BOMBAY
1902
COPYRIGHT, 1900,
BY LONGMANS, GKEEN, AND CO,
ALL RIGHTS RESERVED.
PREFACE.
THIS book contains little more than what is required for
the solution of spherical triangles and related simple practical
problems. The articles on spherical geometry are necessary for
those who have not already studied that subject; for others,
they provide a useful review. More than usual attention has
been given to the measurement of solid angles. The explana-
tions in connection with the astronomical problems are somewhat
fuller than is customary in elementary text-books on spherical
trigonometry.
I am indebted to Mr. W. B. Eite, Ph.B., Fellow in Mathe-
matics at Cornell University, for his kind assistance in reading
the proof-sheets ; and to Mr. A. T. Bruegel, M.M.E., of the
Pratt Institute, Brooklyn, N.Y., for the pleasing character of
the diagrams.
D. A. MURRAY.
CORNELL UNIVERSITY,
May, 1900.
183975
CONTENTS.
CHAPTER I.
REVIEW OF SOLID AND SPHERICAL GEOMETRY.
ART. PAGE
1-4. Planes and lines in space. Diedral angles. Solid angles . . 1
THE SPHERE.
5. The sphere and its plane sections ....... 4
6. Great and small circles on a sphere ....... 4
7. To draw circles about a given pole ....... 6
8-9. Proposition. Problem 7
10. Lines and planes which are tangent to a sphere .... 7
11. Spherical angles 8
ON SPHERICAL TRIANGLES.
12. Definitions 10
13. Propositions 12
14. Correspondence between solid angles and spherical triangles . . 13
15. Propositions ........... 14
16-17. On polar triangles ......... 15
18. Definitions 18
19. Convention 19
20. Shortest line between two points on a sphere 19
PROBLEMS OP CONSTRUCTION.
22. Problems on great circles . . ' . . . . . .21
23-24. Construction of triangles (six cases) ...... 22
CHAPTER II.
RIGHT-ANGLED SPHERICAL TRIANGLES.
25. Spherical Trigonometry 28
2(5. Relations between the sides and angles of a right-angled spherical
triangle. 28
27. On species 32
vii
viii CONTENTS.
ART. PAG*
28. Solution of a right-angled triangle 33
29. The ambiguous case 34
80. Napier's rules of circular parts .35
31. Numerical problems 36
32. Solution of isosceles triangles and quadrantal triangles . . .39
33. Solution of oblique spherical triangles (six cases) .... 39
34. Graphical solution of (oblique and right) spherical triangles . . 42
CHAPTER III.
RELATIONS BETWEEN THE SIDES AND ANGLES OF SPHERICAL TRIANGLES.
36. Derivation of the Law of Sines and the Law of Cosines ... 44
37. Formulas for the half -angles and the half-sides .... 47
38. Napier's Analogies 50
39. Delambre's Analogies or Gauss's Formulas 52
40. Other relations between the parts of a spherical triangle ... 53
CHAPTER IV.
SOLUTION OF TRIANGLES.
41. Cases for solution . . c ....... 54
42. Case I. Given the three sides .... ... 55
43. Case II. Given the three angles 56
44. Case III. Given two sides and their included angle ... 56
45. Case IV. Given one side and its two adjacent angles ... 57
46. Case V. Given two sides and the angle opposite one of them . 57
47. Case VI. Given two angles and the side opposite one of them . 61
48. Subsidiary angles . . 61
CHAPTER V.
CIRCLES CONNECTED WITH SPHERICAL TRIANGLES.
49. The circumscribing circle ......... 62
50. The inscribed circle .... .... 63
51. Escribed circles .......... 64
CHAPTER VI.
AREAS AND VOLUMES CONNECTED WITH SPHERES.
52. Preliminary propositions 66
53. To find the area of a sphere. Area of a zone 67
CONTENTS. ix
ART. PAGE
54. Lunes 69
55. A spherical degree defined . . . . . . . .09
56. Spherical excess of a triangle ........ 70
57. The area of a spherical triangle . . . . . . .71
58. Formulas for the spherical excess of a triangle .... 73
59. The number of spherical degrees in any figure on a sphere. The
spherical excess of a spherical polygon 74
60. Given the area of a figure : to find its spherical excess ... 74
61. The measure of a solid angle 76
62. The volume of a sphere 78
63. Definitions. Spherical pyramid, segment, and sector ... 78
64. Volume of a spherical pyramid ; of a spherical sector ... 79
65. Volume of a spherical segment ....... 80
CHAPTER VII.
PRACTICAL APPLICATIONS.
66. Geographical problem . . . . . , . . . 82
APPLICATIONS TO ASTRONOMY.
68. The celestial sphere 84
69. Points and lines of reference on the celestial sphere ... 86
70. The horizon system : Positions described by altitude and azimuth . 87
71. The equator system : Positions described by declination and hour
angle 88
72. The altitude of the pole is equal to the latitude of the place of
observation ........... 89
73. To determine the time of day ........ 90
74. To find the time of sunrise ........ 91
75. Theorem 92
76. The equator system : Positions described by declination and right
ascension 92
77. The ecliptic system : Positions described by latitude and longitude 93
APPENDIX.
NOTE A. On the fundamental formulas of spherical trigonometry . 95
NOTE B. Derivation of formulas for Spherical Excess .... 98
QUESTIONS AND EXERCISES FOR PRACTICE AND REVIEW . . e 101
ANSWERS TO THE EXAMPLES ......... 113
SPHERICAL TRIGONOMETRY.
CHAPTER I.
REVIEW OF SOLID AND SPHERICAL GEOMETRY.
On beginning the study of spherical trigonometry it is advisable
to recall to mind or learn some of the definitions and propositions
of solid geometry. A clear and vivid conception of the principal
properties of the sphere is especially necessary. The definitions
and theorems which will be used frequently in the following pages,
are quoted in this chapter.*
Planes and Lines in Space. Diedral Angles. Solid Angles.
1. a. Two planes which are not parallel intersect in a straight
line. (Euc. XI. 3.)
6. The angle which one of two planes makes with the other
is called a diedral angle. Thus, in Fig. 1, the two planes BD and
Fia. 1
* As far as possible, references are made to the text of Euclid ; since, of
the numerous geometrical text-books in English-speaking countries, his work
is the one which is most largely used. Those who use a text-book other than
Euclid's can substitute the appropriate references.
1
s/'//r./.'/r.l/-, TRIGONOMETRY. [Cn. I.
. I /•; intersect in the straight line AB, and form the diedral angle
/•'. i no.
c. The planes AE ancM|la7 are called the faces, and the line
. I />' is called the edge, of We diedral angle. The faces are unlim-
ited in extent. The magnitude of the diedral angle depends, not
upon tho extent of its face^rbut only upon their relative position.
(Just as the magnitude or a plane angle depends, not upon the
lengths of its boundary lines, but upon their relative position.)
cZ. If PR be drawn perpendicular to AB in the plane AE, and
PS .be drawn perpendicular to AB in the plane AC, the angle
RPS is called the plane angle of the diedral angle.
e. If a plane is drawn perpendicular to the edge of a diedral
angle, the intersections of this plane with the faces of the diedral
angle form the plane angle of the diedral angle. (See Euc. XL 4.)
Thus, if the plane M be passed through p perpendicular to AB,
the intersections, pr, ps, of the plane M and the planes AE, AC,
form the angle rps which is the plane angle of FABC.
f. All plane angles of the same diedral angle are equal. (See
Euc. XI. 10.) Hence, the plane angle can be taken as the measure
of the diedral angle.
2. a. If a straight line be at right angles to a plane, every
plane which passes through the line is at right angles to that
plane. (Euc. XI. 18.)
b. If two planes which cut one another be each of them per-
pendicular to a third plane, their common section is perpendicular
to the same plane. (Euc. XL 19.)
3. a. When three or more planes meet in a common point,
they are said to form a solid angle, or a polyedral angle, at that
point.
The point in which the planes meet is called the vertex of
the solid angle; the intersections of the planes are called its
edges; the portions of the planes between the edges are called
its faces; the plane angles formed by the edges are called its
face angles; and the diedral angles formed 'at the edges by the
planes are called the diedral angles (or the edge angles) of the
solid angle.
3-4]
GEOMETRY OF THE SPHERE.
Thus, in Fig. 2, for the solid angle formed at 8 : the vertex is
S-, SB, SO, SD, SE, are the edges; BSE, ESD, etc., are the
faces; the face angles are the angles BSE, ESD, DSC, CSB;
the diedral (or edge) angles are BESD, EDSC, etc.
Fio. 2
FIG. 3
b. A solid angle with three faces is called a triedral angle.
Thus, the solid angle at 0 (Fig. 3) is a triedral angle.
(The measurement of solid angles is discussed in Art. 61. The
magnitude of the solid angle in nowise depends upon the lengths
of its edges.)
4. a. The sum of any two face angles of a triedral angle is
greater than the third. (See Euc. XI. 20.)
b. The sum of the face angles of any solid angle is less than
four right angles (Euc. XI. 21). (This is true, in general, only
when the polygon, say BE DC (Fig. 2), formed by the intersec-
tions of the faces with a cutting plane M, does not have a re-
entrant angle ; in other words, when the polygon BE DC is
convex.
Geometry of the Sphere.
For the benefit of those who have not studied the geometry of
the sphere, proofs of a few of its propositions are either out;
lined, or given in detail. Some propositions can be proved very
easily ; hence, only their enunciations are given. Other proper-
ties of the sphere will be proved when they are required. (See
Arts. 53, 54, 57, 62, 65.) The use of a globe on which figures can
be drawn, will be of great assistance to the student. If such a
globe is not at hand, a terrestrial or celestial globe can afford
some service.
4 SPHERICAL TRIGONOMETRY. [Cn. I.
5. The sphere and its plane sections.
a. Definitions. A spherical surface is a surface all points of
which are equidistant from a point called the centre. A sphere is
a solid bounded by a spherical surface. The surface of a sphere
can be generated by the revolution of a semicircle about its
diameter. A radius of a sphere is a straight line joining the
centre to any point on the surface. According to the definition
of a sphere, all the radii of a sphere are equal. A diameter of
a sphere is a straight line passing through the centre and>
terminated at both ends by the surface. A plane section of a
sphere is a figure whose boundary is the intersection of a plane
and the surface of the sphere.
b. Proposition. TJie boundary of every plane section of a sphere
is a circle.
Let the sphere whose centre is at 0 be cut by a plane in the
section ABD ; then ABD is a circle. Through 0 draw OC per-
pendicular to the plane ABD. Let A and B be any two points
in the boundary of the section ABD. Draw
OA, OB, CA, and CB. In the two triangles
OCA and OCB, the angles at C are equal
(both being right angles), the side OC is com-
mon, and the side OA is equal to the side OB,
since both are radii of the sphere. Hence the
triangles are equal in every respect, and CA
is equal to CB. But A and B are any two
points on the boundary of the section ; hence
all points on the boundary are equidistant from C. Therefore
ABD is a circle whose centre is at (7, the foot of the perpen-
dicular let fall from the centre 0 to the cutting plane ABD.
6. Great and small circles on a sphere.
a. Definitions. The section in which a sphere is cut by a plane
is called a Great Circle when the plane passes through the centre
of the sphere ; the section is called a Small Circle when the cut-
ting plane does not pass through the centre of the sphere. Thus,
on a terrestrial globe the meridians and equator are great circles ;
the parallels of latitude are small circles. The Axis of a circle of
5-6.] CIRCLES ON A SPHERE. 5
a sphere is the diameter of the sphere perpendicular to the plane
of the circle ; the extremities of the axis are called the Poles of
the circle and any of its arcs. Thus, in Fig. 4, Art. 5, N and S
are the poles of the circle ABD and of the arcs AB and BD, It
is obvious that all circles made by the intersections of parallel
planes with a sphere have the same axis and poles. For instance,
all parallels of latitude have the same axis and poles, namely,
the polar axis of the earth and the North and South Poles.
6. Propositions relating to great circles.
Every great circle bisects the surface of the sphere; e.g. the
equator bisects the surface of a terrestrial globe.
Any two greatf circles bisect each other; e.g. the meridians
bisect one another at the poles. All great circles of a sphere are
equal ; since their radii are radii of the sphere.
A great circle caTi be passed through any two points on a
sphere ; since a plane can be made to pass through these two
points and the centre of the sphere, and this plane intersects the
surface of a sphere in a great circle. In general, only one great
circle can be drawn through two points on a sphere, since these
points and the centre determine a plane ; but, when the two given
points are at the ends of a diameter an infinite number of great
circles can be drawn through them; e.g. the meridians passing
through the North and South Poles.
c. Definitions. By distance between two points on a sphere is
meant the shorter arc of the great circle passing through them.
It is shown in Art. 20 that this arc is the shortest line that can be
drawn on the surface of the sphere from the one point to the
other. For example, the arc NA in Fig. 4 measures the distance
between the points N and A. [Ex. Distance between N and S ?]
NOTE. The theorem in Art. 20 can be shown mechanically by taking two
points on a parallel of latitude on a globe and letting a string be stretched
taut from one point to the other. The string will not lie on the parallel, but
will evidently be in a plane which passes through the centre of the sphere. If
the two points be on a meridian, the stretched string will lie on the meridian.
By angular distance between two points on a sphere is meant
the angle subtended at the centre of the sphere by the arc joining
the given points. Thus in Fig. 4 the angle NO A is the angular
distance of A from N.
6 srilKRICAL TETGONOMETEY. [On. I.
il. Propositions and definitions relating to small and great circles.
In Fig. 4 till tlin arcs of givat circles, as NA, NB, ND, drawn
from points on the circle ABD to the pole N9 are equal. Thus
the arcs of meridians on a terrestrial globe drawn from a parallel
of latitude to the North Pole are equal. The chords NA, NB,
\I>, are all equal; the angles AON, hON, DON, are likewise
t'qiuil. It thus appears that all points in the circumference of a
circle on a sphere are equally distant from a pole of the circle,
whether the distance be measured by the arc of a great circle
joining one of the points and the pole, or by the straight line
joining the point and the pole, or by the angle which such an arc
or chord subtends at the centre of the sphere.
Definitions. The last mentioned angle is called the angular
radius of the circle. The angular radius of a great circle is evi-
dently a right angle. The polar distance of a circle on a sphere
is its distance from its pole, the distance being measured along
an arc of a great circle passing through the pole. Thus the north
polar distance of a parallel of latitude is its distance from the
North Pole measured along a meridian. The term quadrant,
when used in connection with a sphere, usually means an arc
equal in length to one-fourth of a great circle. The polar dis-
tance of each point on a great circle is evidently a quadrant;
e.g. a point on the equator is at a quadrant's distance from the
North or South Pole. Points on a great circle are equidistant
from both its poles. The polar distance of a circle may be called
the radius of the circle.
' 7. To draw circles upon the surface of a sphere about a given point
as pole.
(a) With a pair of compasses. Open the compasses until the dis-
tance between the points of the compasses is equal to the chord
of the polar distance (or, what is the same thing, the chord sub-
tended by the angular radius) of the required circle. Then, one
point being placed and kept fixed at the pole, the other can describe
the circle.
(&) With a string. Take a string equal in length to the polar
distance of the required circle. If the string be kept stretched
OF
7-10.]
CIRCLES ON A SPHERE.
taut, and one end be fixed at the pole while the other end moves
on the sphere, the required circle will be described..
In order to describe a great circle the polar distance must be
taken equal to a quadrant of the sphere.
8. Proposition. If a point on the surface of a sphere lies at a
quadrant's distance from each of two points, it is the pole of the great
circle passing through these points.
If the point P be at a quadrant's distance
from each of the points A and B, then P is the
pole of the great circle passing through A and
B. Let 0 be the centre of the sphere, and draw
OA, OB, OP. Since PA and PB are quad-
rants, the angles POA and POB are right
angles. Hence PO is perpendicular to the
plane AOB (Euc. XI. 4) ; therefore P is the
pole of the great circle ABL.
9. Problem. Through tivo given points to draw an arc of a great
circle. About each point as a pole draw a great circle (Art. 7).
The two points of intersection of the great circles thus drawn are
each at a quadrant's distance from the two given points; and
hence, by Art. 8, are the poles of the great circle through the two
given points. Accordingly, the required arc will be obtained by
describing a great circle about either of these poles.
NOTE. If the two given points are diametrically opposite, an infinite num-
ber of great circles can be drawn through them. (Art. 6. &.)
10. Lines and planes which are tangent to a sphere.
a. Definitions. A straight line or a plane is said to be tangent
to a sphere when it has but one point in common with the surface
of the sphere. The common point is called the point of contact or
point oftangency.
. 6
8
SPHERICAL TRIGONOMETRY.
[Cn. I.
b. Propositions. (See Fig. 6.)
A plane or a line perpendicular to a radius at its extremity is
tangent to the sphere. [Suggestion for proof : The perpendicular
is the shortest line that can be drawn from a point to a plane.]
A tangent to an arc of a great circle at any point of the arc is
perpendicular to the radius (of the sphere) drawn to the point.
11. On spherical angles.
a. Definitions. The angle made by any two curves meeting in
a common point is the angle formed by the two tangents to the
curves at that point. Thus in Fig. 7,
the angle made by the curves Ci and C2
at the point P, is the angle T^TT* be-
tweeii the tangents to Ci and O2 at P.
(This definition applies to all curves,
whether they are in the same plane or
riot.)
A spherical angle is the angle formed
by two intersecting arcs of great circles
on the surface of a sphere. Thus the angle formed by the
arcs CA and CB (Fig. 8) is a spherical angle. This angle is
the angle ECD between the tangents CE and CD. But ECD is
the plane angle of the diedral angle between the planes CO A and
COB which are the planes of the arcs CA and CB. Thus the
spherical angle is equal to the diedral angle of the planes of the arcs
forming the angle.
C
0i
FIG. 7
FIG. 9
FIG.
b. Propositions. (1) If two arcs of great circles intersect, the
opposite vertical angles thus formed are equal. Thus in Fig. 49,
Art. 57, the angles BAC and B'AC1 are equal.
11.] SPHERICAL ANGLES. 9
(2) If one arc of a great circle meets another arc of a great
circle, the sum of the adjacent spherical angles is equal to two
right angles. Thus in Fig. 49, CAB + CAB' = 2 right angles.
NOTE. It is shown in plane geometry that angles at the centre of a
circle are proportional to their intercepted arcs ; hence, the angles can be
measured by the arcs. Accordingly, if each right angle at the centre of a
circle (Fig. 9) be divided into 90 equal parts called degrees, and the circle be
divided into 360 equal parts, also called degrees, then the number of degrees
(of angle) in any angle AOB is equal to the number of degrees (of arc) in AB,
the arc subtended by AOB. [When it is necessary to distinguish between
dryrc.es of angle and degrees of arc, the former may be called angular
degrees; and the latter arcual degrees.']
c. Proposition. A spherical angle is measured by the arc of a
great circle described with its vertex as a pole and included between
its boundary arcs, produced if necessary :
Let ABC and AB'C be two intersecting arcs of great circles
on the sphere S whose centre is at 0. Pass the plane BOB'
through 0 perpendicular to AC, and let this
plane intersect the planes ABC and AB'C
in the radii OB and OB', and intersect the
sphere in the great circle B'BL. From the
construction, A is the pole of the great
circle B'BL. By Art. 1. e. BOB' is the
plane angle of the diedral angle BACB',
and, accordingly (Art. 11. a), is equal to the
spherical angle BAB'. Now, by the pre-
ceding note, the number of degrees in the
arc BB' is equal to the number of degrees in the angle BOB'.
Hence, the number of degrees in the arc BB' is equal to the num-
ber of degrees in the angle BAB'. In other words, the spherical
angle BAB' is measured by the arc BB' of which A is the pole.
This can be illustrated on a terrestrial globe. For instance, the angle at
the North Pole between the meridians of Paris and New York is 76° 2' 25.5" ;
and this is the number of degrees of arc intercepted by these meridians on
the equator.
d. The great circles drawn through any point on a sphere are
perpendicular to the great circle of which the point is the pole.
10 SPHERICAL TRIGONOMETRY. [Cn. I.
For instance, the meridians of longitude cross the equator at
right angles.
e. The distance of any point on the surface of a sphere, from a
circle traced thereon, is measured by the shorter arc of a great
circle passing through the point and perpendicular to the given
circle ; that is, by the shorter arc of the great circle passing
through the given point and the pole of the given circle. For
example, on a globe the latitude of any place (i.e. its distance in
degrees from the equator) is measured by the arc of the meridian
intercepted between the place and the equator.
N.B. When an arc on a sphere is referred to, an arc of a great circle is
meant, unless expressly stated otherwise.
ON SPHERICAL TRIANGLES.
12. Definitions. A spherical polygon is a portion of the surface
of a sphere bounded by three or more arcs of great circles. The
bounding arcs are the sides of the polygon ;
the points of intersection of the sides are
the vertices of the polygon, and the angles
which the sides make with one another are
the angles of the polygon. A diagonal of
a spherical polygon is an arc of a great
circle joining any two vertices which are
not consecutive.
A spherical triangle is a spherical poly-
FlG n gon of three sides.
Thus, in Fig. 11, ABCD is a spherical
polygon; its sides ate AB, BC, CD, DA', its angles are ABC,
BCD, CD A, DAB; its diagonals are BD and AC; ADC and
ABC are spherical triangles. Since the sides of a spherical
polygon are arcs of great circles, their magnitudes are expressed
in degrees.* The lengths of the sides -can be calculated in terms
of linear units when the radius of the sphere is known.
A spherical triangle is right-angled, oblique, scalene, isosceles, or
equilateral, in the same cases as a plane triangle. The notation
* The reason for expressing the sides of spherical polygons in degrees is
considered more fully in Art. 14.
12.] SPHERICAL TEIANGLES. 11
adopted in discussing the plane triangle will be used for the
spherical triangle ; namely, the triangle will be denoted by ABC,
and the sides opposite the angles A, B, C, will be denoted by
a, b, c, respectively.
Two spherical polygons are equal if they can be applied one to
the other so as to coincide. They are said to be symmetrical when
the sides and angles of the one are respectively equal to the sides
and angles of the other, but arranged in the reverse order.
Thus, the spherical triangles ABC and ^41S1(71 (Fig. 12) are
equal if they can be brought into coincidence, say, by sliding one
of them, as ABC, over the surface of the sphere until it exactly
covers the surface A-^B^C^. Accordingly, it is evident that if
these triangles are equal, the angles A, B, (7, are respectively
equal to the angles Alt Bl} Clt and the sides a, 6, c, are respectively
equal to the sides alt &1? c^* On the other hand, the triangles
ABC and A2B.2C2 are symmetrical if the angles A, B, C, are re-
spectively equal to the angles A2, B2, C2, and the sides a, 6, c, to
the sides a2, 62, c2. In this case, the triangle ABC cannot be
brought into coincidence with A2B2C2 by a sliding motion over
the surface of the sphere.
NOTE 1. Two symmetrical spherical triangles can be brought into coinci-
dence if the surface be covered very thinly with some flexible material. For
then ABC can be lifted up, turned over, and the surface bent (or made to
'spring back') in the opposik direction ; after this treatment, ABC can be
made to coincide with A»BzCz»
NOTE 2. The meaning of the phrase reverse order can be seen clearly
on considering the triangles AiBid and A2B2C2 above. In AiBiCi, on
* Some of the sets of minimum conditions necessary for equality of spheri-
cal triangles are stated in Art. 13.
12 SPHERICAL TRIGONOMETRY. [Cn. I.
going from AI to B\, thence to (7i, and thence to A\, one goes around any
point, within the triangle in a counter-clockwise direction. In AzB^Cz, on
the other liaml, on taking the respective equal angles in the same order as
before, that is, on going from A>2 to B», thence to C->, and thence to A2, one
goes round any point within the triangle A^B^Cz in a clockwise direction.
The directions are indicated by the arrows.
13. Propositions. (1) Two spherical triangles which are on the
same sphere, or on equal spheres, and whose parts are in the same
order (as ABO and A^B^C^ Fig. 12) are equal under the same con-
ditions as plane triangles, viz. :
(a) When two sides and their included angle in the one triangle
are respectively equal to two sides and their included angle in
the other ;
(6) When a side and its two adjacent angles in the one triangle
are respectively equal to a side and its two adjacent angles in the
other ;
(c) When the three sides of the one triangle are respectively
equal to the three sides of the other.
[SUGGESTION FOR PROOFS. Equality can be shown by the same methods
as in plane geometry.]
(2) Two spherical triangles which are on the same sphere, or on
equal spheres, and whose parts are in the reverse order (as ABC and
A2B.2C2, Fig. 12), are symmetrical under the conditions (a), (6), (c)","
above.
[SUGGESTIONS FOR PROOF. Construct* a triangle A\B\C\ which is sym-
metrical to AvBvC*. Under the given conditions, according to the preceding
proposition, ABC and A\BiCi have all their parts respectively equal, and
hence ABC and A^BzCz have all their parts respectively equal, and are
accordingly symmetrical. ]
On a plane two triangles may have three angles of the one
respectively equal to three angles of the other and yet not be
equal. On the other hand, as will be made apparent in Arts.
16,24:
(3) On the same sphere, or on equal spheres, two triangles which
have three angles of the one respectively equal to three angles of the
other, are either equal or symmetrical.
* For the construction of spherical triangles under various conditions, see
Art. 24.
13-14.] TRIEDRAL ANGLES: SPHERICAL TRIANGLES. 13
14. Correspondence between the face angles and the diedral angles
of a triedral angle on the one hand, and the sides and angles of a
spherical triangle on the other.
B'
FiO. 13
Take any triedral angle 0-A'B!Cf; let a sphere of any radius,
OA say, be described about 0 as centre ; and let the intersections
of this sphere with the faces OA'B', OB'C', and OC'A', be the
arcs AB, BC, and CA respectively. The sides of the spherical
triangle ABC, namely, AB, BC, CA, measure the face angles,
AOB, BOG, COA, of the solid angle 0-A'B'C' (Art. 11. b, Note).
By Art. 11 the angles CAB, ABC, BCA, of the spherical triangle
ABC are the diedral angles between the planes of the sides, that
is, the diedral angles of the solid angle 0-A'B'C'.
Hence, to find the relations existing between the face angles
and the edge angles of a triedral angle, is the same thing as to
find the relations between the sides and angles of the spherical
triangle, intercepted by the faces, upon the surface of any sphere
whose centre is at the vertex of the triedral angle.
NOTE 1. The number of degrees in the intercepted arcs does not depend
upon the radius of the sphere. Thus, in Fig. 13, if a sphere is described
with a radius OAi, about 0 as a centre, the number of degrees in the inter-
cepted arc AiBi is the same as the number of degrees in the intercepted arc
AB, for each number is the same as the number of degrees in the angle
A' OB'.
Since the face angles and diedral angles of a triedral angle are not altered
by varying the radius of the sphere, the relations between the sides and
angles of the corresponding spherical triangle are independent of the length
of the radius.
14 SPHERICAL TRIGONOMETRY. [Cn. I.
NOTE 2. Since the side of a spherical triangle measures the angle sub-
tended by it at the centre, the side is measured in degrees or radians. (See
Art. 12.) By " sin AB," for example, is meant the sine of the angle AOB,
subtended by AB at the centre 0.
NOTE 3. A three-sided spherical figure, one or more of whose sides is not
an arc of a great circle, is not regarded as a spherical triangle. For exam-
ple, the figure bounded by an arc of a parallel of latitude and the arcs of two
meridians does not correspond to a triedral angle at the centre of the sphere,
and is not a spherical triangle as defined in Art. 12.
NOTE 4. A triedral angle, and its corresponding spherical triangle, can be
easily constructed. From stiff cardboard cut out a circular sector having
any arc between 0° and 360°. On this sector draw any two radii, taking
care, however, that no one of the three sectors thus formed shall be greater
than the sum of the other two. Along these radii cut the cardboard partly
through. Bend the two outer sectors over until their edges meet ; a figure
like 0-ABC (Fig. 13) will be obtained. (Find what happens if the above
precaution in drawing the radii is not taken.)
This perfect correspondence between the sides and angles of a
spherical triangle on the one hand, and the face angles and die-
dral angles of the solid angle subtended at the centre of the
sphere by the triangle on the other hand, is very important, both
for the deduction of the relations between these sides and angles
and for the solution of practical problems. This correspondence
holds in the case of any spherical polygon and the solid angle
subtended by it at the centre of the sphere. (The student may
inspect Fig. 11.) Hence, from any property of polyedral angles an
analogous property of spherical polygons can be inferred, and vice
versa.
15. Propositions. (1) Any side of a spherical triangle is less
than the sum of the other tivo sides. This follows from Arts. 14
and 4. a.
COR. Any side of a spherical polygon is less than the sum of
the remaining sides.
(2) The sum of the sides of a spherical polygon (not re-entrant)
is less than 360°. In other words : Tlie perimeter of any (non-re-
entrant) spherical polygon is less than the length of a great circle.
This important proposition follows from Arts. 14 and 4. b.
15-16.] POLAR TRIANGLES. 15
(3) In an isosceles spherical triangle the angles opposite the
equal sides are equal.
(4) .The arc of a great circle drawn from the vertex of an
isosceles spherical triangle to the middle of the base is perpen-
dicular to the base, and bisects the vertical angle.
(5) If two angles of a spherical triangle are unequal, the oppo-
site sides are unequal, and the greater side is opposite the greater
angle.
COR. If two edge angles of a triedral angle are unequal, the
opposite face angles are unequal, and the greater face angle is
opposite the greater diedral angle.
(6) If two sides of a spherical triangle are unequal, the oppo-
site angles are unequal, and the greater angle is opposite the
greater side.
Ex. Give the corresponding proposition for a triedral angle.
Propositions (3)— (6) can be proved in the same way as the cor-
responding propositions in plane geometry.
ON POLAR TRIANGLES.
16. a. NOTE. Three straight lines on a plane, no two of which are
parallel, intersect in three points, and form one triangle. Three great circles
on a sphere have six points of intersection, and form eight spherical triangles.
Thus, on a globe, the equator and any two great circles through the poles
have as intersections the two poles and the four points where the two great
circles cross the equator ; and there are eight triangles formed, namely, four
in the northern hemisphere and four in the southern.
6. Definitions. If great circles be described with the vertices
of a spherical triangle, say ABC (Fig. 14), as poles ; and if there
be taken that intersection of the circles described with B and C
as poles which lies on the same side of BC as does A, namely A^\
and similarly for the other intersections ; then a spherical triangle
is formed, which is called the polar triangle of the first tri-
angle ABC.
Two spherical polygons are mutually equilateral when the sides
of the one are respectively equal to the sides of the other,
whether taken in the same or in the reverse order ; the polygons
16
N ."///: /,' 1C A L 77,' / G ONOMETE Y.
[Cn. I.
are mutually equiangular when the angles of the one are respec-
tivcly equal to the angles of the other, whether taken in the same
<>r in the reverse order.
B
FIG. u
c. Proposition. If the first of two spherical triangles is the polar
triangle of the second, then the second is the polar triangle of the first.
If A'B'C' (Fig. 14) is the polar triangle of ABC, then ABC is
the polar triangle of A'B'C'. Since A is the pole of the arc
B'C', the point A is a quadrant's distance from B'. Also, since C
is the pole. of B'A, the point C is a quadrant's distance from B'.
Since B' is thus a quadrant's distance from both A and C, it is
the pole of the arc AC (Art. 8). Similarly it can be shown that
A' is the pole of the arc BC, and that C' is the pole of the arc
AB. Hence ABC is the polar triangle of A'B'C',
d. Proposition. In two polar triangles, each angle of the one is
the supplement of the side opposite to it in the other.
Let ABC and A'pO (Fig. 15) be a
pair of polar triangles, in which A, B, C,
A, B'} C', are the angles, and a, b, c,
a', I', c', are the sides.' Then
A = 180 - a', A = 180 - a,
B = 180 - b', B1 = 180 - &,
C = 180 - c', C' = 180 - c.
Produce the arcs AB and AC to meet
B'C' in L and M respectively.
(6-17.] SUM OF ANGLES OF A TR] ANGLE. 17
Since B' is the pole of ACM, B'M = 90° ; and
since C' is the pole of ABL, LC' = 90°.
Hence B'M + LC' = 180° ;
that is, B'M + MC'+LM = 1SQ°,
or jB'C" + 7v^=180°. (1)
Since A is the pole of the arc B'C', the arc LM measures the
angle A (Art. 11. c).
Hence, (1) becomes A + a' = 180°, or A = 180° -a'.
The other relations can be proved in a similar manner.
COR. If two spherical triangles are mutually equiangular,
their polar triangles are mutually equilateral. If two spherical
triangles are mutually equilateral, their polar triangles are
mutually equiangular.
NOTE. On account of the properties in (cZ), a triangle and its polar are
sometimes called supplemental triangles.
e. The use of the polar triangle. Because of the fact that the sides and
angles of a triangle are respectively supplementary to the angles and sides of
its polar triangle, many relations can be easily derived by reference to the
polar triangle. For, if a relation is true for spherical triangles in general,
then it is true for the polar of any triangle. Let the relation be stated for the
polar triangle ; in this statement express the values of the sides and angles of
the polar triangle in terms of the angles and sides of the original triangle ; the
statement thus derived expresses a new relation between the parts of the
original triangle. This will be exemplified in later articles.
17. Proposition. The sum of the angles of a spherical triangle is
greater than two, and less than six, right angles.
'4 Let ABC be any spherical triangle ; it is required to show that
Construct the polar triangle A'B'C'. Then, by Art. 16. d,
A -f a' = 180°, B + V = 180°, C + c' = 180°.
18 SPHERICAL TRIGONOMETRY. [Cn. I.
Hence, on adding, A + B + C+a' + &' +c' = 540°,
or, .-1 + 1* + Cy = 540° - (a' + 6' + c').
Now [Art. 15 (2)] a' + &' + c' is less than 360°, and. greater
than 0°.
.-. (A + B + C)= 540° - (something less than 360° and greater
than 0°).
... A + B + C > 540° - 360°, i.e. A+B + C> 180° ;
and A + B+ C< 540° - 0°, i.e. A + B+C< 540°.
18. Definitions, a. The amount by which the sum of the
three angles of a spherical triangle is greater than 180° is called
its spherical excess. It is shown in Art. 57 that the area of a
triangle depends upon its' spherical excess.
6. A spherical triangle may have two right angles, three right
angles, two obtuse angles, or three obtuse angles. For example,
on a globe the spherical triangle bounded by any arc (not 90°) on
the equator and the arcs of the meridians joining the extremities
of the former arc to the North Pole, has two right angles ; if the
arc on the equator is a quadrant, the triangle • has three right
angles. The polar of the triangle whose sides are 35°, 25°, 15°,
has three obtuse angles. A spherical triangle having two right
angles is called a bi-rectangular triangle, and a spherical triangle
having three right angles is called a tri-rectangular triangle. A
triangle having one side equal to a quadrant is called a quadrantal
triangle; one having two sides each a quadrant is said to be
bi~quadrantal, and one having each of its three sides equal to a
quadrant is said to be tri-quadrantal.
c. A lune is a spherical surface bounded by the halves of two
great circles. The angle of the lune is the angle made by the two
great circles. For instance, on a globe the surface between the
meridians 10° W. and 40° W. is a lune ; the angle of this lime is
equal to 30°. On the same circle or on equal circles lunes having
equal angles are equal. (For they can evidently be made to
coincide.)
18-20.]
SHORTEST LINE ON A CIRCLE.
19
FIG. 16
19. On the convention that each side of a spherical triangle be less
than 180°» in spherical geometry and trigonometry it is found convenient
to restrict attention to triangles the sides of which
are each less than a semicircle or 180°. (This con-
vention can be set aside when it is necessary to con-
sider what is called the general spherical triangle,
in which an element may have any value from 0°
to 360°.) A triangle such as ADBC (Fig. 16) which
has a side ADB greater than 180°, need not be con-
sidered ; for its parts can be immediately deduced
from the parts of ACS, each of whose sides is less
than 180°. It is easily proved that if an angle of a
spherical triangle is greater than 180°, the opposite
side is also greater than 180°, and vice versa. Thus,
in the triangle ADBC, if the angle ACB is greater than 180°, so is the side
ADB ; and it ADB is greater than 180°, so is the opposite angle. [Sugges-
tion for proof : Produce the arc AC to meet the arc ADB.~]
20. Proposition. The shortest line that can be drawn on the sur-
face of a sphere between two given points is the arc of a great circle,
not greater than a semicircle, which joins the points.
Let A and B be any two points on a sphere, and let ACB be a
great-circle arc not greater than a semicircle; then ACB is the
shortest line that can be drawn from A to B on the sphere.
About A as a pole describe a circle DCE
with radius AC, and about B as a pole
describe a circle FCG with radius BC. It
will be shown (1) that C is the only point
which is common to both these circles;
(2) that the shortest line that can be drawn
from A to B on the surface must pass
through C.
(1) Take any point G, other than C, on
the circle FCG. Draw the great-circle arcs
AEG and BG. By Art. 15 (1),
AG+GB>AB; i.e. AG + GB > AC + CB.
Now AE = AC, and GB = CB.
Hence AE + GB = AC + CB ;
and, accordingly, AG > AE.
Fio. 17
20 SPHERICAL TRIGONOMETRY. [Cn. I.
Therefore G is outside of the circle DCE. But G is any point
(other than Gy) on the circle FOG. Hence C is the only point
common to the circles DCE and FOG.
(2) Let ADFB be any line drawn on the surface from A to
J3, but not passing through C. Whatever the character of the
line AD may be, a line exactly like it can be drawn from A to (7;
and a line like BF can be drawn from B to C.
[This can be seen by regarding A-DCE as a cap fitting closely to the
sphere, and supposing that this cap revolves about A until D is at C. Then
a line exactly like AD is drawn from A to C.]
These lines being drawn, there will be a line from A to B
which is less than ADFB by the part DF. It has thus been
proved that a line can be drawn from A to B through C which is
shorter than any other line from A to B which does not pass
through C. But C is any point on the great-circle arc from A to
B. Hence the shortest line from A to B must pass through every
point in ACB, and, accordingly, must be the arc ACB itself.
NOTE. This proposition can also be proved by the method of limits. It is
shown that the length of any arc on a sphere is equal to the limit of the
sum of the lengths of an infinite number of infinitesimal great-circle arcs
inscribed in the given arc. (See Rouche" et De Comberousse, Traite de
Geometric.) See Art. 6. c.
PROBLEMS OF CONSTRUCTION.
21. The actual making of the following constructions will add
much to the clearness and vividness of the notions of most stu-
dents about the surface of a sphere. An easy familiarity with
the problems of Arts. 23, 24, which discuss the construction of
triangles, will place the student in an advantageous position with
respect to spherical trigonometry. This position is similar to
that occupied by him, through his knowledge of the construction
of plane triangles, when he entered upon the study of plane
trigonometry. (See Plane Trigonometry, p. 20, Note, Art. 21,
Art. 34 (to Case I.), Art. 53.)
N.B. The student should try to make these constructions for himself,
and should fall back upon the book only as a last resort.
21-22.] PROBLEMS ON GREAT CIRCLES. 21
22. Problems on great circles.
(1) To find the poles of a given great circle. About any two
points of the given circle as poles, describe great circles ; their
intersections will be the poles required (Art. 8).
(2) To draw a great circle through two given points. About the
two given points as poles, describe great circles ; about either of
the intersections of these circles as a pole, describe a great circle ;
this will be the circle required. (See Arts. 8, 9.)
(3) To cut from a great circle an arc n° long. Separate the
points of the compasses by a distance equal to a chord which
subtends a central angle of n° in a circle whose radius is equal to
the radius of the sphere ; place the points of the compass on the
great circle ; the intercepted arc will be the one required.
(A) To draiv a great circle through a given point perpendicular to
a given great circle. Find a pole of the given circle by (1) ; draw
a great circle through this pole and the given point by '(2) ; this
circle will be the one required (Art. 11. d).
(5) To construct a great circle making a given angle with a given
great circle, the point of intersection being given. About the given
point of intersection as pole, describe a great circle 5 on this circle
lay off an arc, measured from the given circle, having as many
(arcual) degrees as there are (angular) degrees in the given angle ;
draw a great circle through the extremity of this arc and the given
point of intersection; this will be the circle required (Art. 11. c).
(6) To construct a great circle passing through a given point^and
making a given angle with a given great circle. [When the given
point is on the given circle this problem reduces to problem (o).]
It is easily shown that the angle betweenxtwo great circles is
equal to the angular distance (Art. 6. c) between their poles.
Hence, find a pole of the given circle by (1) ; about this point as
pole describe a second circle whose angular radius (Art. 6. d) is
equal to the given angle ; the pole of the required circle must be
on this second circle. About the given point as pole describe a
great circle ; if the required problem is possible, this circle will
either touch or intersect the second circle. The points of contact
or intersection are the poles of two great circles, each of which
will satisfy the given conditions.
Ex. Discuss the case in which the given point is the pole of the given circle.
22 SPHERICAL TRIGONOMETRY. [Cn. I.
23. Construction of triangles. The three sides and the three
angles of a spherical triangle constitute its six jmrts or elements.
If any three of these six parts be known, the triangle can be con-
structed. The construction belongs to geometry; the computation
of the three remaining parts, when three parts are given, is an
important part of spherical trigonometry. The sets of three
parts that can be taken from the six parts of a spherical triangle
are as follows :
I. Three sides.
II. Three angles.
III. Two sides and their included angle.
IV. One side and the two adjacent angles.
V. Two sides and the angle opposite one of them.
VI. Two angles and the side opposite one of them.
NOTE. There are four construction problems in the case of plane triangles
(Plane Trig., Art. 53). When three angles of a spherical triangle are given,
there is only one spherical triangle (with the triangle symmetrical to it), as
will presently appear, which satisfies the given conditions. When three
angles of a plane triangle are given, there is an infinite number of triangles,
of the same shape, but of different magnitudes, which have angles equal to
the three given angles. Cases IV. and VI. above, in which two angles are
given, reduce to a single case in plane trigonometry, namely, the case in
which one side and two angles are given ; since the sum of the three angles
of any plane triangle is 180°.
24. To construct a spherical triangle.
I. Given the three sides. On any great circle lay off an arc
equal to one of the given sides [Art. 22 (3)]. About one extrem-
ity of this arc as pole, describe a circle with a radius (arcual)
equal to the second of the given sides ; about the other extremity
of the arc as pole, describe a circle with a radius equal to the
third of the given sides. By arcs of great circles join either of
the points of intersection of the last two circles to the extremities
of the arc first laid off; the triangle thus formed satisfies the
given conditions.
Ex. 1. Compare the construction in the corresponding case in plane
triangles.
23-24.] CONSTRUCTION OF TRIANGLES. 23
Ex. 2. How many triangles are possible when the first arc is laid off ?
Are these triangles equal or symmetrical ?
Ex. 3. Construct ABC: (a) Given a = 70°, b = 65°, c = 40°; (6) Given
a = 120°, b = 115°, c = 80°.
Ex. 4. Determine approximately the angles of these triangles. (See
Arts. 11. c, 34.)
II. Given the three angles. Calculate the sides of the polar
triangle (Art. 16. d); construct it by I. above; construct its
polar (Art. 16. 6) j the latter triangle is the one required.
Ex. 1. How many triangles can be drawn when one side of the polar
triangle is fixed ? Are these triangles equal or symmetrical ?
Ex. 2. Discuss the corresponding case in plane triangles.
Ex. 3. Construct ABC: (a) Given A = 85°, B = 75°, C = 55° ; (6) Given
A = 75°, B = 105°, C = 100°.
Ex. 4. Determine approximately the sides of these triangles.
III. Given two sides and their included angle. Take any point
on any great circle; through this point draw a circle making
with the first circle an angle equal to the given included angle
[Art. 22 (5)]; from the chosen point and on the first circle
bounding this angle, lay off an arc equal to one of the given sides ;
from the same point and on the second circle bounding the angle,
lay off an arc equal to the other given side. Join the extremities
of these arcs by the arc of a great circle; the triangle thus
formed is the one required.
Ex. 1. How many triangles can be made when the first circle and the
point are chosen ? Are these possible triangles equal or symmetrical ?
Ex. 2. Discuss the corresponding case in plane triangles.
Ex. 3. Construct ABC : (a) Given a = 75°, 6 = 120°, C = 66° ; (6) Given
b = 35°, c = 70°, A = 110°.
Ex. 4. Determine approximately the remaining parts of these triangles.
IV. Given a side and its two adjacent angles.
Either : a. On any arc of a great circle lay off an arc equal to
the given side ; its extremities will be taken as two vertices of
the required triangle. Through one extremity of this arc draw a
great circle making with the arc an angle equal to one of the
given angles; through the other extremity of the arc draw a
24 SPHERICAL TRIGONOMETRY. [Cn. I.
great circle making with the arc (and on the same side as the
angle first constructed) an angle equal to the other of the given
angles. The point of intersection of these two circles which is on
the same side of the arc as the two angles, is the third vertex of
the required triangle.
Or: b. Calculate the corresponding parts of the polar tri-
angle ; construct it by III. ; construct its polar ; this will be the
triangle required.
Ex. 1. How many triangles are possible when the first arc is chosen ?
Are these triangles equal or symmetrical ?
Ex. 2. Discuss the corresponding case in plane triangles.
Ex. 3. Solve Problem III. by means of IV. a, and the polar triangle.
Ex. 4. Construct ABC : (a) Given a = 75°, B = 65°, C = 110° ; (6) Given
-6 = 110°, A = 40°, C = 63°.
Ex. 5. Determine approximately the remaining parts of these triangles.
V. Given two sides and the angle opposite to one of them. [First,
review the corresponding case in plane geometry.]
To construct a triangle ABC when a, 6, A, are known : Through
any point A of a great circle ALA' A draw the semicircle, ACA',
making an angle A LAC equal to the given angle A. From this
semicircle cut off an arc AC equal to b. About C as a pole, with
an arc equal (in degrees) to the side a, describe a circle. The
intersection B of this circle with ALA1 will be the third vertex of
the required triangle, A and C being the other two vertices.
Four cases arise, viz. : —
(1) When the circle described about C fails to intersect ALA' ;
(2) When it just reaches to ALA' ;
(3) When it intersects the semicircle ALA' in but one point ;
(4) When it intersects the semicircle ALA> in two points.
Case (1) is represented in Figs. 18, 22 ; case (2), in Figs. 10, 23 ; case (3),
in Figs. 20, 24 ; and case (4), in Figs. 21, 25. In Figs. 18 and 22 the angle
A is respectively acute and obtuse ; and similarly for each of the other pairs
of figures.
NOTE. In Figs. 18-25 AKA' is a great circle in the plane of the paper,
and ALA' A is a great circle at right angles to that plane, ALA1 being above
the surface of the paper, and the dotted A A' being below. In Figs. 18-21,
24.]
CONSTRUCTION OF TRIANGLES.
25
Fia. 18
Fia. 19
Fia. 32
Fia. 31
FIG. 35
26 SPHERICAL TRIGONOMETRY. [Cn. I.
angle A is acute [equal to PAK(W°}-KAC~], and the arc AC is in front
of the paper. In Figs. 22-25, angle A is obtuse [equal to PAK(9Q°) -f
1\A C), and the arc AC is behind the paper.
In Fig. 21 there are two triangles (not equal or symmetrical) that satisfy
the given conditions; amHikewisemn'Fig. 25. Hence V. is an ambiguous
case in spherical geometry.
In each figure let the perpendicular arc CP be drawn from C to the semi-
circle ALA', and let its length be called^. [See Ex. 1, p. 101.]
A. When angle A is acute :
Fig. 18 shows that the triangle required is impossible, if CB < CP, i.e. if
a<p.
Fig. 19 shows that the triangle required is right angled if CB = CP, i.e.
v ifa=.p.
Fig. 20 shows that there is but one triangle which satisfies the given con-
ditions, if
CB> CP, CB>CA, and CB< CA' }
i.e. if a >_p, a > b, and a < 180° — b.
^ Similarly, there is only one triangle if a >p, a<b, and a > 180° — b.
Fig. 21 shows that there are two triangles which satisfy the given condi-
tions, if
CB> CP, CB< CA, and CB < CA' ;
4 i.e. if a >p, a<b, and a < 180° - b.
B. When angle A is obtuse :
Fig. 22 shows that the triangle required is impossible, if CGB > CGP, i.e.
v \ia>p.
Fig. 23 shows that the triangle required is right angled, if CGB = CGP,
i.e. if a — p.
Fig. 24: shows that there is but one triangle which satisfies the given con-
ditions, if
CLB < CGP, CLB < CA and CLB > CA' ;
J i.e. if a < p, a < b, and a > 180° - 6.
Similarly, there is only one triangle if a <p, a > b, and a < 180° — b.
Fig. 25 shows that there are two triangles which satisfy the given condi-
tions, if
CLB< CGP, CLB> CA, and CLB> CA' ;
V i.e. if a <p, a > b, and a > 180° - 6.
In Fig. 25 produce PGC to meet the great circle ALA A in P'. Then
CP1 = 180° . — p. Since AC and CA ' are each greater than CP', it follows
J that a > 180° -p.
24.] CONSTRUCTION OF TRIANGLES. 27
It is also apparent from Figs. 20 and 21 that the triangle is impossible,
if A is acute, a > &, and a > 180° - b ;
and it is apparent from Figs. 24 and 25 that the triangle is impossible,
if A is obtuse, a < &, and a < 180° - 5.
Some special cases which may be investigated by the student,
are indicated in the exercises on this chapter at page 101.
VI. Given two angles and the side opposite one of them. Calculate
the corresponding parts of the polar triangle ; construct it by V. ;
construct its polar ; this is the required triangle. There may be
two solutions, since the triangle first constructed may have two
solutions.
Ex. 1. Construct ABC: (a) Given a = 52°, 6 = 71°, A = 46°; (6) Given
a = 99°, 6 = 64°, A = 95°.
Ex. 2. Construct ABC : (a) Given A = 46°, B = 36°, a = 42° ; (6) Given
A = 36°, B = 46°, a = 42°.
Ex. 3. Determine approximately the remaining parts of these triangles.
N.B. Questions and exercises on Chapter I. will be found at pages
101-102.
CHAPTER II.
RIGHT-ANGLED SPHERICAL TRIANGLES.
25. Spherical trigonometry. Spherical trigonometry treats of
the relations between the six parts of a triedral angle, or, what is
the same thing (Art. 14), the relations between the six parts of the
corresponding spherical triangle intercepted 011 the surface of the
sphere. In Art. 24 it has been seen how a triangle can be con-
structed when any three parts are given ; Chapters II. and III.
are concerned with showing how the remaining parts can be
computed when any three parts are known. In this chapter the
relations between the sides and angles of a right-angled spherical
triangle are deduced.1* Throughout the book the relations between
trigonometric ratios, discussed in plane trigonometry, will be
employed.
26. Relations between the sides and angles of a right-angled spheri-
cal triangle.
Case I. The sides about the right angle both less than 90°.
Let ABC be a spherical triangle which is right angled at C and
on a sphere whose centre is at 0. First suppose that a and b are
B
FIG. 26
* These relations can also be obtained from the relations, derived in
Chapter III., between the parts of any spherical triangle or triedral angle.
28
25-26.] RELATIONS BETWEEN SIDES AND ANGLES. 29
each less than 90°. (It is easily shown, geometrically, that c is
then less than 90°.) Draw OA, OB, OC. Take any point P on
OA ; in the plane OA C draw PL at right angles to OA, and let
it meet OC in L ; in the plane OAB draw PM at right angles to
OA, and let it meet OB in M ; and draw ML. Since PL and
PM are perpendicular to the line OA, the line OA is perpendicu-
lar to the plane LPM (Euc. XI. 4) ; and, therefore, the plane
LPM i£ perpendicular to the plane OAC (Euc. XI. 18). Also,
the plane gCB is perpendicular to the plane OAC, since C is a
right angle. Hence, LM, the intersection of the planes LPM
and OCB, is perpendicular to the plane (MO (Euc. XI. 19) ; and
hence, MLP and MLO are right angles.
In the triangle 0PM, the angle 0PM being right,
OM OL OM
Now, angle POM = c, ^ = cos POL = cos 6,
OL
and — = cos LOM = cos a.
OM
.-. cos c = cos a cos 6. (1)
In the triangle LPA/, angle PLM= 90°, and angle LPM = A;
LM
LM OM sin LOM
OM
(2)
sine
PL
PL OP tan POL
OP
Also,
whence, cos ^ = (3)
tanc
SPHERICAL TRIGONOMETRY. [CH. II.
LM
LM OL
Also, tan LPM = - = -
OL
whence, tan A = (4)
sin 6
On remarking that A, a, denote an angle and its opposite, side,
and that b denotes the other side, the relations for angle B cor-
responding to (2), (3), (4), can be written immediately ; viz. :
(2'); cos7J = - (3<); tanS = (4').
sin c tan c sin a
These relations for B can also be deduced directly, by taking
any point on OB and making a construction similar to that shown
in Fig. 26.
Moreover,
sin A = tan A cos A = • - = • -— [By (3), (4).]
smi tanc tanc cos*
or -i i • T-» COS A
Similarly. sin B = — — • (o)
cos a
Also, cos c = cos a cos b = ~ -^— • - — • [By (1), (o), (o').]
Sill -Z5 S1U -*T-L
.\ cos c = cot A cot B. (6)
For convenience of reference, relations (l)-(6) may be grouped
together :
cos c - cos a cos b. (1)
elii A — **tn<^. gin ^ — Sin ., /0\ /of\
sin c sin c
tanf ' (3>> (3')
sTn^' W, W
sin B = 55Lfl. /5x /5»x
cos a
cos c = cot A cot JS. (6)
26.]
RELATIONS BETWEEN SIDES AND ANGLES.
81
Case II. The sides about the right angle both greater than 90°.
. In Fig. 27, C denotes the right angle, and the sides a, b, are
each greater than a quadrant.
Form the lime CO by producing the sides a and b to meet in
C'.. Then ABC' is a right triangle in which the sides about the
right angle are each less than 90°.
.-. cos c = cos BO cos AC' = cos (180 - a) cos (180 - b).
Hence
Also,
cos
whence,
tanAB
cos c =^cos aVos b.
cos (180° - BAG) =
tan b
tan
cos ^4 =
tan c
In a similar manner the other relations in (l)-(6) can be shown
to be true for ABC (Fig. 27).
NOTE. ABC' is said to be co-lunar with ABC. Every triangle has three
co-lunar triangles, one corresponding to each angle.
Case III. One of the sides about the right angle less than 90°,
and the other side greater than 90°.
c
FIG. 28
In ACB let C = 90°, a > 90°, and b < 90°. Complete the lune
BB'. Then ACB' is a right-angled triangle in which the sides
about the right angle are each less than 90°.
.32 SPHERICAL TRIGONOMETRY. [On. II.
In ACS', cos AB' = cos AC cos CB' ;
i.e. cos (180° - c) = cos b cos (180° - a) ;
whence cos c = cos a cos fr.
Again,
cos CAB1 = tan^C; i.e. cos (180° - BAG) = tan^C
tan^LB'' tan (180° - BA) '
tan 6
whence. cos A =
tan c
In a similar way the other relations in (1)— (6) can be shown
to be true for ABC (Fig. 28).
erJV 27. On species. Two parts of a spherical triangle are said to
be of the same species (or of the same affection) when both are less
than 90°, both greater than 90°, or both equal to 90°. Formula
(1), Art. 26, shows that the kifflojgniw of a right-angled spherical
triangle is less than 90° when the sides about the right angle are
both greater or both less than 90° ; and it shows that the hypotenuse
is greater than 90° when the sides are of different species. Formulas
(4) and (4') show that in a right-angled spherical triangle (since
the sines of the sides are positive) an angle and its opposite side
are of the same species. These important properties can also be
deduced geometrically.
EXAMPLES.
N. B. It is advisable to remember the result of Ex. 1.
1. State relations (l)-(6), Art. 26, in words.
(1). cos hyp. = product of cosines of sides.
(6). cos hyp. = product of cotangents of angles.
(2), (2'). sin angle = sin opposite side ~ sin, hyp.
(3), (3'). cos angle = tan adjacent side -f- tan hyp.
(4), (4'). tan angle = tan opposite side -=- sin adjacent side.
(5), (5'). cos angle = cos opposite side x sin other angle.
[Compare (2), (3), (4), with the corresponding formulas in plane tri-
angles.]
2. Deduce formulas (l)-(4) by means of a figure in which P is anywhere
on OB (see Fig. 26).
27-28.] SOLUTION OF A RIGHT-ANGLED TRIANGLE. 33
3. Deduce formulas (l)-(4) by means of a figure in which P (see Fig. 26)
is : (a) in OA produced ; (6) in OB produced ; (c) at the point A ; (d) at
the point B.
4. The two sides about the right angle of a spherical triangle are 60° and
75° ; find the hypotenuse and the other angles by means of relations (1), (4),
(4'), Art. 26. Check (or test) the result by means of other formulas.
5. In ABC, given ^1 = 47° 30', .5 = 53° 40', C=90°; find the remaining
parts, and check the results.
6. Solve some of the examples in Art. 31, and check the results.
28. Solution of a right-angled triangle.
N.B. The student is advised to investigate this subject independently ;
and, before reading this article, to put in writing in an orderly manner his
ideas about the solution of right triangles. These ideas will thus be made
clearer in his mind, and his subsequent reading will be easier.
In a right triangle there are five elements beside the right
angle. These five elements can be taken in groups of three in
ten different ways. Each of these ten groups is involved in the
ten relations derived in Art. 26; the three elements of each
group are, accordingly, connected by one relation.
Ex. (a) Write all the groups of three that can be formed from a, 6, c,
A, B, such as a, &, c ; a, b, A ; etc.
(6) Write the relation connecting the elements of each group.
It follows that if any two elements of a right-angled spherical
triangle besides the right angle be given, then the remaining three
elements can be determined. The method of finding a third
element is as follows :
Write the relation involving the two given elements and the re-
quired element; solve this equation for the required element.
Check (or test). When the required elements are obtained, the
results can be checked by examining whether they satisfy rela-
tions which have not been employed in the solution, and, pref-
erably, the relation involving the newly found elements.
E.g., suppose that A, &, are known, C being 90° ; then a, c, B,
are required. Side a can be found by (4) ; side c, by (3) ; and
angle B, by (5). The values found for a, c, B, can be checked
by (31)-
84 SPHERICAL TRIGONOMETRY. [Cn. IL
NOTE 1. The cosine, tangent, and co-tangent of sides and angles greater
than 90°, are negative. Careful attention must be paid to the algebraic signs
of the trigonometric functions appearing in the work.
NOTE 2. The properties stated in Art. 27 are very useful.
NOTE 3. Determine each element from the given elements alone. If an
element is found erroneously and then used in rinding a second element, the
second element will also be wrong.
The ten possible groups of three elements correspond to the
following six cases for solution, in which the given elements are
respectively :
(1) Two sides. (4) Two angles.
(2) Hypotenuse and a side. * (5) Side and opposite angle.
(3) Hypotenuse and an angle. (6) Side and adjacent angle.
Before proceeding to the solution of numerical examples, it is
necessary to refer more particularly to one of these cases ; and
also to call attention to the fact that the ten formulas for right
triangles (Art. 26) may be grouped in two very simple and con-
venient rules.
j/* 29. The ambiguous case. When the given parts are a side and
its opposite angle, there are two triangles which satisfy the given
conditions. For, in ABC (Fig. 29), let C = 90°, and let A and CB
(equal to a) be the given parts. Then, on completing the lune
AA1, it is evident that the triangle A'BC also satisfies the given
conditions, since BOA' = 90°, A' = A, and CB = a. The remain-
o C
Fia. 29
ing parts in A'BC are respectively supplementary to the remain-
ing parts in ABC', thus A'B = 180°-c, A'C=18Q°-b, A'BC
= 180° - ABC.
This ambiguity is also apparent from the relations (l)-(6),
Art. 26. For, if a, A, are given, then the remaining parts, namely,
29-30.] NAPIER1 S RULES. 35
c, b, B} are all /determined from their sines [see (2), (4), (5'),] ;
and, accordingly, c, b, B, may each be less or greater than 90°.
On the otMr hand, if, for instance, a and c are given, then b is
determined from its cosine by (1) ; and there is no ambiguity,
because b is less or greater than 90° according as cos b is respec-
tively positive or negative.
N.H. Both solutions should be given in the ambiguous case, unless some
information is given which serves to indicate the particular solution that is
suitable.
30. Napier's rules of circular parts. The ten relations derived in
Art. 26 are all included in two statements, which are called Napier'' s rules of
circular parts, after the man who first announced them, Napier, the inventor
of logarithms.
Let ABC be a triangle right-angled at C. Either draw a right-angled
triangle, and mark the sides and angles as in Fig. 31, or draw a circle and
mark successive arcs as in Fig. 32, in which b, a, Co-B, Co-c, Co-A, are
CO-B c°;9— --
/ \ / No
Co-c
C f
\ .<-' \ ^
Co-A wv*
Pia. 31 Fia. 33
arranged in the same order as 5, a, B, c, A, in Fig. 30. (Here Co-B, Co-c,
Co-A, denote the complements of B, c, and A, respectively.) The five
quantities, a, b, Co-B, Co-c, Co-A, are known as circular parts. That is, the
right angle being omitted, the two sides and the complements of the hypote-
nuse and the other ang'.es are called the circular parts of the triangle.
In Figs. 31 and 32 each part has two circular parts adjacent to it, and two
circular parts opposite to it. Thus, on taking a, for instance, the adjacent
parts are b, Co-B, and the opposite parts are Co-c, Co-A. If any three parts
be taken, one of them is midway between the other two, and the latter are
either its two adjacent parts or its two opposite parts. Thus, if a, b, Co-A,
be taken, then b is the middle part, and a, Co-A, are the adjacent parts ; if a,
b, Co-c, be taken, then Co-c is the middle part, and a, b, are opposite parts.
Ex. Take each of the circular parts in turn, write its opposite parts and
adjacent paxts, thus getting ten sets in all.
T
36 SPHERICAL TRIGONOMETRY. [Cn. II
Napier"1 s rules ofafacular parts are as follows :
I, The sine of me middle part is equal to the product of the tangents oj
the adjacent parts. kr
II. The sine of tltg middle part is equal to the product of th'e cosines of
the opposite parts.
(The I's, a's, and o's are lettered thus, in order to aid the memory.)
These rules are easily verified. For example, on taking a for the middle
part,
sin a = tan b tan (90° - J3) = tan b cot B. [See Art. 26 (4').]
sin a = cos (90° - A) cos (90° - c) = sin J.,sin c. [See Art. 26 (2).]
Again, on taking Co-A for the middle part,
sin (90° - A) - tan 6 tan (90° - c), i.e. cos A = tan b cot c.
[See Art. 26 (3).]
sin (90° — A)= cos a cos (90° — _B), i.e. cos A = cos a sin B.
[See Art. 26 (5').]
In a similar way each of the remaining three parts can be taken in turn
for the middle part, and the remaining six relations of Art. 26 shown to
agree with Napier's rules.*
NOTE. One may either memorize the relations in Art. 26 (or Ex. 1,
Art. 27), or use Napier's rules.- Opinions differ as to which is the better
thing to do.
Ex. 1. Verify Napier's rules by showing that they include the 10 relations
in Art. 26.
Ex. 2. Prove Napier's rules.
31. Numerical problems. In solving right triangles the pro-
cedure is as follows :
(1) Indicate the two given parts and the three required parts.
* This is only a verification of Napier's rules. One proof of the rules
would consist of the derivation of the relations in Art. 26 plus this verifica-
tion. These rules were first published by Napier in his work Mirifici
Logarithmorum Canonis Descriptio in 1614. Napier indicated a geometrical
method of proof, and deduced the rules as special applications of a more
general proposition. They are something more than mere technical aids to
the memory. For an explanation of this and of their wider geometrical
interpretation, see Charles Hutton, Course in Mathematics (edited by T. S.
Davies, London, 1843), Vol. II. pp. 24-26; Todhunter, Spherical Trigo-
nometry, Art. 68; E. O. Lovett, Note on Napier's Rules of Circular Parts
(Bulletin Amer. Math. Soc., 2d Series, Vol. IV. No. 10, July, 1898).
31.]
NUMERICAL PROBLEMS.
37
(2) Write the relations -that will be employed in the solution,
and note carefully the algebraic signs of the functions involved.
The noting of these signs will serve to show (unless a part is
determined from its sine) whether a required part is less or greater
than 90°.
(3) For use as a check, write the relation involving the three
required parts.
(4) Arrange the work as neatly and clearly as possible.
N.B. Attention may be directed to the notes in Art. 28. Also see Plane
Trigonometry, Art. 27 (particularly p. 45, notes 2, 4-6), and Art. 59, p. 107.
NOTE. The trigonometric function of any angle can be expressed in
terms of some trigonometric function of an angle less than 90°. See Plane
Trigonometry, Art. 45.
EXAMPLES.
1. Solve the triangle ABC, given :
C = 90°, Solution * : c =
a = 44° 30', A =
b = 71° 40' B =
Formulas : cos c = cos a cos b,
tan A = tan a -f- sin b,
tan B = tan b -^ sin a.
Check : cos c = cot A cot B.
Logarithmic formulas :
[If necessary ; see PL
Trig., Art. 27, Note 6.]
log sin a = 9.
log cos a = 9.85324 - 10
log tan a - 9.99242 - 10
log sin b = 9.97738 - 10
log cos & = 9.49768- 10
log tan b = 0.47969
Check :
log cos c = log cos a + log cos 6,
log tan A = log tan a — log sin 6,
log tan B = log tan b — log sin a,
log cos c = log cot A + log cot B (check).
^
-10
log cos c = 9.35092 - 10
logtanJ. = 0.01504
log tan B = 0.63403
.'. c = 77° 2'.1.
= 76° 55'. 5.
.-. log cot .4 = 9. 98497 -10
log cot B = 9.36595 - 10
.-. log cos c = 9.35092 - 10
* To be filled in.
•• u -
IE;T£
38
SPHERICAL TRIGONOMETRY.
[Cn. II.
NOTE. Spherical triangles, like plane triangles, can also be solved
without the use of logarithms. (See Plane Trigonometry, examples in
Arts. 27, 55-62.)
2. Solve the triangle ABC, given :
(7=90°,
.4 = 57° 40',
« = 48°30'.
Formulas
sine =
sin B —
sin a
sin A
tan a
tan A
cos A
Solution : c =
b =
B =
Check : sin B =
sin b
sin c
cos a
log sin a = 9.87446 - 10
log cos a = 9.82126 - 10
log tan a = 0.05319
log sin A = 9.92683 - 10
log cos ^4 = 9. 72823 -10
log tan A = 0.19860
The check gives log sin B = 9.90696.
log sin c =9.94763-10
log sin b =9.85459 - 10
log sin B = 9. 90697 - 10
* /. c = 62° 25'.4, or 117° 34'.6.
6 = 45°40'.9, or 134° 19'. 1.
JB = 53°49'.3, or 126° 10'. 7.
On combining the results according to the principles of Art. 27, the
solutions are :
(1) c = 62°25'.4, 6= 45°40'.9, B= 53°49'.3;
(2) c=117°34'.6, 6 = 134°19'.l, B = 126° 10'. 7.
3. Solve Ex. 1 without using logarithms.
- 4. Given c = 90°, A = 57° 40', a = 108° 30'. Show both by geometry and
trigonometry why the solution is impossible.
5. Solve the triangle ABC, in which C = 90°, and check the results, given :
. (1) a = 36° 25' 30", b = 85° 40' ; (2) c = 120° 20' 30", a = 47° 30' 40" 'j
v
(3) c = 78° 25', A = 36° 42' 30"
(5) a = 76° 48', A = 82° 36' ;
(7) a = 47° 40', A = 30° 43' ;
(9) a = 108° 45', J? = 37°42';
(11) A = 110° 27', B = 74° 36' ;
(13) ^1 = 124° 30', b = 25° 40' ;
(4) A = 63° 18', B = 37°J'
b = 39° 50i 20", A - 47°
P = TTTWSO" B = 80° 40* 20"
B = 47° 50' ;
(10) c= 78° 20',
a = 108°
(14) c = 84° 47',
: *
= 39° 43'.
7, d 2.. rj -/d
32^-33.] OBLIQUE TRIANGLES. 39
32. Solution of isosceles triangles and quadrantal triangles.
Isosceles Triangles. Plane isosceles triangles can be solved by
means of right triangles, as shown in Plane Trigonometry, Art. 32.
A spherical isosceles triangle can be solved in a similar way,
namely, by dividing it into two right triangles by an arc drawn
from the vertex at right angles to the base. This arc bisects the
base and the vertical angle.
Quadrantal Triangles. The polar triangle of a quadrantal tri-
angle (Art. 18) is right-angled (Art. 16. d). Hence, a quadrantal
triangle may be solved by solving its polar triangle by Arts. 28,
31, and then computing the required parts of the quadrantal
triangle by Art. 16. d.
EXAMPLES.
1. Solve the triangle ABC, in which A and B are equal, and check the
results, given :
(1) a = 54° 20', c = 72° 54' ; (2) a = 66° 29', A = 50° 17' ;
(3) a = 54° 30', (7=71°; (4) c = 156°40', (7=187° 46'.
2. Solve the trianplft..^ Kf7, given ;
j>) c = 90°, (7 = 67° 12', a^EL
(2) c = 90°, A = 136° 40', B = 105° 47'.
33.* Solution of oblique spherical triangles. It has been seen
(Plane Trigonometry, Art. 34) that oblique plane triangles can be
solved by means of right triangles. Oblique spherical triangles
can also be solved by means of right spherical triangles. Relat-
ing to spherical triangles there are six problems of computation;
these correspond to the six problems of construction discussed in
Arts. 23, 24. If any three parts of a triangle are given, the tri-
angle can be constructed and the remaining parts can be com-
puted. The several cases for computation will now be solved
with the help of right-angled triangles, f
(In the figures in this article the given parts are indicated by crosses.)
N.B. The student is advised to try to solve Cases II.-VI. before reading
the text.
* When time is limited this article may be omitted, or merely glanced over.
t Other methods of solving triangles are shown in Chap. IV.
40
SPHERICAL TRIGONOMETRY.
[Cn. II.
Case I. Given the three sides. In ABC (Figs. 33, 34) let a, 6, c,
be given, and A, B, C, be required. From C draw CD at right
angles to AB, or AB produced. Let the segments AD and DB be
denoted by m and n, respectively. If the direction from A to B
is taken as the positive direction along the arc AB, then m is
positive in Fig. 33 and negative in Fig. 34, while n is positive in
both figures.
C C
FIG. 33 Fio. 34
Special formula. In each figure,
cos a — cos n cosp, and cos b = cos m cos p.
cos a cos b
cosp =
cos n cos m
. cosn _cosa
cosm cos b
cos n — cos m cos a — cos
[Composition and division.]
cos n + cos m cos a + cos b
From this, on applying Plane Trigonometry, Art. 52, formulas (7), (8),
tan i (n + m) tan | (n - m) = tan | (a + 6) tan | (a - 6). (1)
Now n + m = c] hence, n — m can be found by (1). Then the
segments m and n can each be determined. The triangles ADC
and jBZ><7 can then be solved by Arts. 28, 31 ; and the solution of
ABC can be obtained therefrom.
Ex. 1. Solve Exs. 1, 2, Art. 42, by the method outlined above.
Ex. 2. Show how to solve this case when the perpendicular is drawn
from A to BC.
33.]
OBLIQUE TRIANGLES.
41
Case II. Given the three angles. Solve the polar triangle by
the method used in Case I. ; and therefrom (Art. 16. d) compute
the parts of the original triangle.
Ex. Solve Exs. 1, 2, Art. 43, by this method.
Case III. Given two sides and their included angle. In ABC
(Fig. 35) let a, c, B, be given. Draw AD at right angles to BC,
or BC produced.
In ABD, c and B are known ; hence,
BAD} AD, and BD can be found. In
ADC, AD and DC (equal to a — BD)
are now known ; hence DAC, ACD, and
AC can be found. Also, CAB = CAD
+ DAB. The student can examine the
case in which AD falls outside ABC.
Ex. 1. Show how to solve the triangle by
drawing a perpendicular arc from C to AB.
Ex. 2. Solve Exs. 1, 2, Art. 44, by means
of right triangles.
y — -"^
Two meth-
Case IT. Given a side and the two adjacent angles.
ods of solution may be employed.
Either: (1) Solve the polar triangle by the method used in
Case III. ; and therefrom compute the parts of the original
triangle.
Or: (2) In ABC (Fig. 36) let A, B, c, be given. Draw the arc
BD at right angles to AC. In ADB,
AD, DB, and ABD can be found, since
A and AB are known. Now DBC =
ABC -ABD. In DBC, DB and DBC
are now known ; hence" BC, CD, and C
can be found. Then AC = AD + DC.
The student can examine the case in
which BD falls outside ABC.
Ex. 1. Solve the triangle by drawing a dif-
ferent perpendicular.
Ex. 2. How may solution (2) aid in the solution of Case III. ?
Ex. 3. Solve Exs. 1, 2, Art. 45, by means of right triangles.
42
SPHERICAL TRIGONOMETRY.
[Cn. II.
Case V. Given two sides and the angle opposite one of them. (Tliis
may be an ambiguous case; see Art. 24, V.)
In ABC (Fig. 37) let a, c, A, be given. From B draw the arc
BD at right angles to AC to meet AC or
AC produced. In ABD, c and A are
known ; hence AD, DB, and ABD can be
found. In DBC, DB and a are now
known; hence DBG, C, and DC can be
found. Then AC = AD + DC, and
= ABD +
Ex. 1. Examine the cases in which BD falls
outside ABC.
Ex. 2. Examine the case in which two triangles satisfy the given con-
ditions.
Ex. 3. Solve Exs. 1, 2, Art. 46, by means of right triangles.
Case VI. Given two angles and the side opposite one of them.
Like Case V. this may be ambiguous; see Art. 24, VI. Two
methods of solution may be employed.
Either: (1) Solve the polar triangle by the method used in
Case V. ; and therefrom compute the parts of the original triangle.
Or: (2) In ABC (Fig. 38) let A, C, c, be known. From B
draw BD at right angles to AC to meet AC, or AC pro-
duced. Solve the triangle ABD; then
solve the triangle DBO. The parts of
ABC can be computed from these solu-
tions.
Ex. 1. How may (2) aid in the solution of
Case V. ?
Ex. 2. Solve Exs. 1, 2, Art. 47, by means of
right triangles.
Ex. 3. Solve the numerical examples in
Art. 24.
34. Graphical solution of (oblique and right) spherical triangles.
A plane triangle can be solved graphically by drawing to scale
a triangle that satisfies the given conditions, and then measuring
the required parts directly from the figure (Plane Trigonometry,
34.] GRAPHICAL SOLUTION. 48
Arts. 10, 21). A spherical triangle can be solved graphically by
drawing (Art. 24) upon any sphere a triangle that satisfies the
given conditions, and then measuring the required parts of the
triangle. The sides and angles (see Art. 11. c) can be measured
with a thin, flexible, brass ruler, on which a length equal to a
quadrant or a semicircle of the sphere is graduated from 0° to
90° or 180° respectively.
Small slated globes can be obtained fitting into hemispherical cups, whose
rims are graduated from 0° to 180° in both directions. With such a globe,
cup, and a pair of compasses, the constructions discussed in Art. 24 and the
measurements referred to in this article are easily made.
If the student has the means at hand, it is advisable for him to
solve some of the numerical problems graphically.
N.B. Questions and exercises on Chapter II. will be found at p. 102.
CHAPTER III.
RELATIONS BETWEEN THE SIDES AND ANGLES OF
SPHERICAL TRIANGLES.
i.
35. In this' chapter some relations between the sides and
angles of any spherical triangle (whether right-angled or oblique)
will be derived. In the next chapter these relations will be used
in the solution of practical numerical problems. The first two
general relations (namely, the Law of Sines and the Law of
Cosines), which are by far the most important, can be derived in
various ways. In a short course it may be best to deduce these
laws by means of the properties of right-angled triangles as set
forth in Art. 26; and, accordingly, this method is adopted here.
These laws are also derived directly from geometry in Note A at
the end of the book. It may be stated here that the geometrical
derivation will strengthen the student's understanding of the
subject, and will show more clearly the correspondence (Art. 14)
between the parts of a spherical triangle and the parts of a
triedral angle.
36. The Law of Sines and the Law of Cosines deduced by means
of the relations of right-angled triangles.
A. Derivation of the Law of Sines.
Let ABC (Figs. 39, 40) be any spherical triangle. From B
draw the arc BD at right angles to AC to meet AC, or AC pro-
duced, in D.
In ABD, sin BD — sin c sin A ;
in CBD, sin BD = sin a. sin C (Fig. 39)
= sin a sin BCD (Fig. 40) = sin a sin BCA.
Hence, in both figures, sin a sin C = sin c sin A.
sin a __ sin c
sin A sin C
44
35-36.]
LAW OF COSINES.
45
Similarly, by drawing an arc from C at right angles to AB, it
sin a sin b
oan be shown that
sin A sin B
. sin a _ sin b
sin A sinJ5
sine
sin C
V
(1)
In words : In a spherical triangle the sines of the sides are pro-
portional to the sines of the opposite angles. (Compare Plane
Trigonometry, Art. 54, I.)
B. Derivation of the Law of Cosines.
cos BC = cos CD cos DB
= cos '(6 - AD) cos DB, or cos (AD - b) cos DB
= cos b cos AD cos DB -f sin b sin J.Z) cos DB.
But f>na Xl /•) r>r»« D72 — nne /^ • ft A- -J , $/2>r
(a)
and
cos ^4£> cos DB = cos c ; <txr<
sin .4Z) cos D£ = £2iCjm^D = cos c tan ^
IX
(2)
= cos c tan ccosA= sin c cos A. ' ^
Hence, on substituting in (a),
cos a = cos b cos c -f- sin b sin c cos A .
Similarly, or by taking the sides in turn,
cos b = cos c cos a + sin c sin a cos B,
cos c — cos a cos b + sin a sin b cos C.
In words : In a spherical triangle the cosine of any side is equal
to the product of the cosines of the other two sides plus the product of
the sines of these two sides and the cosine of their included angle.
(Compare Plane Trigonometry, Art. 54, II.)
46 SPHERICAL TRIGONOMETRY. [Cn. III.
NOTE i. The law of cosines, (2), is the fundamental and the most
important relation in spherical trigonometry. For, as shown in Note A,
it can be deduced directly ; the law of sines, (1), can be deduced from it ;
all other relations follow from these ; and the relations for right triangles,
Art. 26, can be deduced from the relations for triangles in general, on letting
C be a right angle. The formulas for cos a, cos 6, cos c, were known to the
Arabian astronomer Al Battani in the ninth century. (See Plane Trigo-
nometry, p. 166.)
C. The Law of Cosines for the angles. Relation (2) holds for
all triangles, and, accordingly, for A'B'C', the polar triangle qf
ABC. (See Fig. 14, Art. 16.) That is,
cos a' = cos b' cos c' + sin b' sin c' cos A9.
.-. cos (180° -^A) = cos (180° - B) cos (180° - C)
+ sin (180° - J3) sin (180° - C) cos (180° - a). [Art. 16. d.]
.*. — cos A. = (— cos B) (— cos (7) -f- sin J5 sin (7(— cos a).
.*. cos A = - cos B cos C + sin B sin C cos a. \/ (3)
Similarly, cos B = — cos C cos A -f- sin C sin A cos 6,
cos C = — cos A cos B -f sin A sin B cos c.
Relation (3) can also be derived by means of right-angled tri-
angles.
NOTE 2. From (2), cos A = COS a ~ COS b COS C.
sin 6 sin c
h The denominator in the second member is always positive. If ^differs
I more from 90° than does 6, then cos a is numerically greater than cos 6, and,
/accordingly, greater than cos b cose; hence cos A and cos a have the same
/ sign, and thus, a and A are in the same quadrant.
Similarly, a and A are in the same quadrant when a differs more from
V^gO^than does c.
~~From (3), hTwhich cos a = cos A + cos B cos C,
sin B sin C
it can be shown in a similar way that if A differs more from 90° than does
B or (7, then a and A are in the same quadrant.
Ex. 1. Derive cos 6 and cos c by means of right triangles.
Ex. 2. Derive cos A and cos B by means of right triangles.
Ex. 3. Derive cos C from cos c by means of the polar triangle.
37.] tffc THE HALF-ANGLES. 47
37. Formulas for the half-angles and the half-^Hfc
[Compare the method and results of this article with those of
Art. 62, Plane Trigonometry'].
I. The half-angles.
-™ A j o/» /o\ A cos a — cos b cos c /-, N
From Art. 36. (2), cos A = — — (1)
sin b sin c
cos a — cos b cos c
/. 1 — cos A = L
sin b sin c
_ cos b cos c + sin b sin c — cos a
sin 6 sin c
_ cos (6 — c) — cos a
sin 6 sin c
. o sin2 i ^ = 2 sin J- (a — & -f c) sin j- (a + 6 — c) t
sin 6 sin c
[Ptae Trigonometry, Art. 52, (8).]
On putting a + b-\-c = 2s, then — a + b -f- c = 2 (s — a),
a — b + c = 2 (s — fr), and a + b — c = 2 (s — c).
sin2 1 A — s^n (g ~ ^ s*n (8 ~ CX
sin b sin c
Similarly, from (1),
l + cos^^l + cosffi-cos6cosc
sin & sin c
'.
_ cos a + sin b sin c — cos b cos c
sin V6 sin c
_ cos a — cos (6 + c)_
sin b sin c
/. 2 cos2 ! ^1 = 2 sin
sin 6 sin c
.'. cos2 J ^ = gin* sin (^ -a). (3)
sin 6 sin c
48
SPHERIC A L TRIGONOMETE Y.
[Cu. III.
and hence, tan * A = J»in(* -») sin (*-<£.
* sin s sm (s — a)
(4)
Therefore, tan^ = * /sin(«-a)sin(8-6)siD^--c)t
sin(s— a) * sins
hence, if tan r = ->m (• - «) sin (8 - 6) sin (g - c) ,
sin s
then tan \ A =
Like
written
namely,
formulas can be similarly derived for J J5 and £ (7; or they may be
immediately on observing the symmetry in formulas (4) and (5);
_ /
'
tan A B =
sin a sin c
sing sin(g — ft)
sin a sin c
in (g — a) sin (g— c)
sin g sin (g— ft)
tanr
=A/
COS £
sin(g — a)sin(g— ft)
sin a sin ft
= /sing sin (g - c)
sin a sin b
tan } (7 --JsinQ-a)sinQ-
sin s sin (s — c)
tanr
sin (s- ft)
sin (s — c)
It is shown in Art. 50 that r is the radius of the circle inscribed
in the triangle ABC. Article 50 may be read at this time.
NOTE. By geometry, 2 g < 360° and ft + c > a. Hence, g — a is positive
and less than 180°. Similarly, g — 6, g — c, are positive. Therefore, the
quantities under the radical signs are positive. The positive signs must be
given to each radical, for ^4, B, C, are each less than 180°, and, consequently,
IA,1B,IC, are each between 0° and 90°.
EXAMPLES.
1. Derive each of the above formulas.
2. Given a = 58°, b = 80°, c = 96°. Find A, B, C.
3. Given a = 46° 30', b = 62° 40', c = 83° 20-. Find A, B, C.
The results in Exs. 2, 3, may be checked by Art. 36, (1).
37.] THE HALF-SIDES. 49
II. The half-sides. From Art. 36, (3),
cos ^ + cos -B CQS ff
cos a —
sin 5 sin (7
On finding 1 — cos a and 1 + cos a, combining and simplifying
in the manner followed for the half-angles, and putting
the following formulas are obtained :
sin J a =
cos a =
"
tan -- a =
2
Let
'cos(S-
an l\ cos
Similarly, or from (8) and (9) by symmetry,
then tan | a - tan R cos (8 — A).
(8)
(9)
sin A sin C ' sin A
sin ^1 sin C ' sin A sin I?
Ian a & -A008 # CM (0-.g anir -A -<*»
- -
tan I 6 - tan U cos (5f - B) , tan | c = tan JB cos (« - C) . (11)
It is shown in Art. 49 that R is the radius of the circumscribing
circle of the triangle ABC. Article 49 may be read at this time.
NOTE 1. Formulas (8)-(ll) can also be derived from formulas (4)-(7)
by making use of the polar triangle, as done in Art. 36. C.
Xl^oxE 2. Since A+B+C lies between 180° and 540° (Art. 17), S lies
/between 90° and 270° ; hence, cos S is negative, and, accordingly, — cos S
I is positive. Since all the other functions under the radical signs are positive,
V^e whole expression under each radical sign is positive.
NOTE 3. The positive value of the radical is taken, since each side
(Art. 19) is less than 180°
50 SPHERICAL TRIGONOMETRY. [Cu. III.
EXAMPLES.
1. Derive formulas (10) from the values of cos b and cose.
2. Derive formulas (10) from formulas (6) by means of the polar trial
3. In ABC, given A = 78° 40', B = 63° 50', C = 46° 20'. Fin
[SUGGESTION. Either use formulas (8)-(10); or, solve the polar triangle,
and thence obtain the parts of the original triangle. [The results may be
checked by using both these methods, or by Art. 36, (1).]
4. In ABC, given A = 121°, B = 102°, C = 68°. Find a, b, c.
5. Show that cos (S — A) is positive.
38. Napier's Analogies. On dividing tan | A by tan | B (Art. 37),
there is obtained,
tan i A _ sin (s — 6)
tan i B ~~ sin (s — a)
From this, by composition and division,
tan A A -f tan | B _ sin (s — b) + sin (s — a)
tan \A — tan 1 B ~ sin (s — b) — sin (s — a) *
This, by Plane Trigonometry, Arts. 44. B, 52 (also, see Art. 61),
reduces to
sin i A cos i J3 + cos i ^ sin ^ ^ _ 2 sin 1 (2 s — a — b) cos i (a — b) ^
sin ±Acos±B — cos ^ J. sin 1 J3 ~~ 2 cos % (2 s — a — 6) sin 1 (a — 6) '
sin^ + B) tan|c
and thence, to — ? - = - ? -- (i)
sinl(^-B) tani(a-6)
On multiplying tan ^ J. by tan | -B, there is obtained
sn s
i e sin ± A sin 1 1? _ sin (8 — c)
cos ^ A cos | ^ sin s
From this, by composition and division,
cos \ A cos ^ B — sin i ^4 sin i B sin s — sin (s — c)
cos ±Acos±B+ sin ±Asin^B~ sin s + sin (s — c)
2 cos |- (2 s — c) sin i c
~~ 2 sin i (2 s — c) cos £ c
cosJ-Oi + B) tan|c
Whence, - = - ? - . (2)
38.] NAPIER'S ANALOGIES. 51
Either, on proceeding in a similar way with tan \ a and tan \ b
[Art. 37, (8), (10)], or, on applying (1) and (2) to the polar tri-
angle, there is obtained,
sin - (a + 6) cot £ C
(3)
sin * (a -6) tan | (^1-1?)
cosi(a + 6) cot|C
and - = — p- — (4)
cos I (a — 0) tan | (.4 + 1?)
Relations (l)-(4) are known as Napier 's Analogies.*
NQTE 1. Compare (3) with formula (2) Art. 61, Plane Trigonometry.
/NOTE 2. The numerators in (3) are always positive, since a+ 6 + c < 360°
y&nd C< 180°. It follows, accordingly, that a - b and A- B must have the
I same sign. This also follows from the geometrical fact [Art. 15, (5)] that
/ the greater angle is opposite the greater side.
NOTE 3. In relation (4), cot \ C and cos \(a — 6) are positive quantities ;
/ hence cos£(a + 6) and tan %(A + B} have the same sign ; and, accordingly,
X^&(g^+ 6) and %(A -f jB) are of the same species (Art. 27).
NOTE 4. Derivation of (3) by applying (1) to the polar triangle. On
applying (1) to the polar triangle A'B'C' (Fig. 14, Art. 16),
sin \ (A* + B') _ tan ^ c'
sin \(A' - B'} ~ tan £(«' - b')'
tan £(180° - C)
^
sin $(180° - a - 180° - 6) tan $(180° - A - 180° - B) '
- £ a + fr) _ tan(90° - £ O)
Whence
sin J(6 - a) tan $(B - A)
5) _
sin £(«-&) tan i (^1-5)
NOTE 5. For a geometrical deduction of Napier's Analogies and the for-
mulas in Art. 39, see M'Clelland and Preston, Treatise on Spherical Trig-
onometry, Part L, Art. 56, and the article Trigonometry in the Encyclopedia
Britannica (9th edition).
* That is, Napier's proportions. For a long time the word analogy was
used in English in one of its original Greek meanings, namely, a proportion
(i.e. an equality of ratios). This use of the word is now obsolete, and is
only retained in a few phrases such as the above. Napier (see Art. 30, and
Plane Trigonometry, Art. 1) discovered these proportions and gave them
in his work, Mirifici logarithmorum canonis descriptio, in 1614
52 SPHERICAL TRIGONOMETRY. [Cn. III.
EXAMPLES.
1. Express Napier's Analogies in words.
2. Write the analogies involving B and C, A and C, b and c, a and c.
3. Derive some of the analogies in Ex. 2.
). Delambre's Analogies or Gauss's Formulas.
/* By Plane Trigonometry, Art. 46, (1),
sin %(A + .B) = sin £ A cos £ 1? + cos £ J. sin £
By Art. 37, (4), (6), -
^-8ipO»-*OcoslC,
sin a sin 6 sin c
and p.™ j ^ «« i p - sin (g ~ «) sin ^ sin (* ~ «) - sm (s - a) CQC ^
sin c ' sin a sin 6 sin c
sin c
«-o-6)cos|(o-6)
In a similar way it may be shown that
!,) = «»*£ -»«»IC, (2)
B) = sinja (4)
Formulas (l)-(4) are known as Delambre's Analogies, and also as Gauss's
Formulas or Equations.*
* These formulas were discovered by Karl Friedrich Gauss (1777-1855),
one of the greatest of German mathematicians and astronomers, and pub-
lished without proof in his Theoria Motus Corporum Ccelestium in 1809 ;
thus they bear his name. They were, however, published earlier by Karl
Brandon Mollweide of Leipzig (1774-1825) in Zach's Monatliche Correspon-
dcnz for November, 1808. They were earliest discovered by Jean Baptiste
Joseph Delambre (1749-1822), a great French astronomer, in 1807, and pub-
lished in the Connaissance des Temps in 1808. The geometrical proof (see
Note 5, Art. 38) was the one originally given by Delambre. This proof was
rediscovered and announced by M. W. Crofton in 1869, and published in the
Proceedings of the London Math. Soc., Vol. III. (1869-1871), p. 13.
39-40.] DELAMBRE'S ANALOGIES. 53
NOTE 1. Equations (3) and (4) can also be derived by applying (1) and
(2) to the polar triangle. >-«VT
NOTE 2. Delarabre's Analogies can also be deduced by help of Napier's
Analogies. (See Todhunter, Spherical Trigonometry, Art. 54 ; Nature, Vol.
XL. (1889, Oct. 31), p. 644.)
NOTE 3. On the other hand, Napier's Analogies can be easily derived
from Delambre's Analogies ; namely, on dividing corresponding members,
one by the other, in (1) and (3), (2) and (4), (4) and (3), (2) and (1).
EXAMPLES,
1. Write Delambre's Analogies involving B and C, and G and A.
2. Derive (3) and (4) from (1) and (2), using the polar triangle.
3. Derive Delambre's Analogies from Napier's Analogies.
4. Derive some of the analogies in Ex. 1 directly.
40. Other relations between the parts of a spherical triangle. The
preceding articles of this Chapter present few more relations than
are required for the solution of spherical triangles. Between the
parts of a spherical triangle there are many other relations which
are interesting and useful for many purposes, and which either
set forth, or lead to the discovery -of, important geometrical prop-
erties * of spherical triangles. t
For example, if in equation (2) Art. 36, the value of cos c in the second
equation that follows, be substituted, then
cos a = cos a cos2 b + sin a sin b- cos b cos C -f sin b sin c cos A ;
•whence, on putting for cos2 b its value 1 — sin2 6, dividing by sin 6, and
transposing, it follows that
cos a sin b — sin a cos b cos C = sin c cos A.
Eive similar relations can be derived by permuting the letters ; and on
applying these six relations to the polar triangle, six others can be derived.
To pursue this topic further is beyond the scope of this book,
which aims to give little more than the simplest elements of
spherical trigonometry and what is absolutely required for the
solution of spherical triangles. Those who are interested can
refer to the works on. spherical trigonometry by M'Clelland and
Preston (Macmillan & Co.), Casey (Longmans, Green, & Co.),
Bowser (D. C. Heath & Co.), and others.
N.B. Questions and exercises on Chapter III. will be found on page 104.
* — ,
* Instances in which geometrical properties are deduced by means of
trigonometry, are given in Art. 27, Art. 36, (Note 2), Art. 38, (Notes 2, 3).
CHAPTER IV.
SOLUTION OF TRIANGLES.
N.B. The student is recommended to work one or two examples in each
set in this chapter before reading any of the text.
41. Cases for solution. This chapter is concerned with the
numerical solution of spherical triangles. In all there are six
cases for solution ; these correspond respectively to the six cases
for construction which were discussed in Arts. 23, 24. In these
cases the given parts are as follows :
I. Three sides.
II. Three angles.
III. Two sides and their included angle.
IV. One side and its two adjacent angles.
V. Two sides and the angle opposite one of them.
VI. Two angles and the side opposite one of them.
With slight changes the procedure described in Art. 31 is rec-
ommended. Figures may be helpful. Of formulas adapted for
logarithmic computation, the necessary ones are (1) Art. 36, (4)-
(11) Art. 37, and (l)-(4) Art. 38. If the polar triangle is used
in finding the solution, then I. and II. constitute one case, like-
wise III. and IV., and likewise V. and VI. ; and the necessary
formulas are (1) Art. 36 (4)-(7) or (8)-(ll) Art. 37, and (1), (2),
or (3), (4) Art. 38. Cases V. and VI. must be examined as to
ambiguity; and accordingly, they give more trouble than the
others. Unless the triangle satisfies the conditions specified in Arts
15, 17, 24 V., its solution is impossible.
Checks. The results obtained should always be checked. Delam-
bre's Analogies and formulas which have not been used in the
course of the solution, may be used as check formulas.
N.B. Before doing any of the numerical work the student should try to
get a clear idea of the figure of the triangle upon a sphere. This geometrical
54
41-42.] NUMERICAL SOLUTIONS. 55
conception will enable him to make a reasonable estimate of what the results
will be; this estimate will help him to detect wild results that may be
obtained by making numerical errors. For example, in ABC let a = 110°,
6 = 114°, C= 10° ; and suppose that the result c= 76° presents itself. A person
who has drawn a figure of the triangle on a sphere, or one who has geometri-
cal imagination sufficient to give him an idea of the look of the given triangle,
will at once see that the result, c = 76°, must be wrong. In working spherical
triangles it is much better not to proceed blindly.
42. Case I. Given the three sides.
EXAMPLES.
1. In ABC, given a = 47° 30', b = 55° 40', c = 60° 10'. Find A, B, C.
Jttnmdat: ^nr^1™^^^
sin t
tan r tan r
tan C=
, , .
sin(s-a) sin (a -6) sin (s - c)
Check : Law of Sines, or Napier's Analogies, or Delambre's Analogies.
Logarithmic formulas :
log tan2 r - log sin (s - a} + log sin (s - 6) + log sin (s - c) - log sin s, etc.
Check : log sin a — log sin A = log sin b — log sin B = log sin c — log sin C.
a = 47° 30' log sin s = 9.99539 - 10 .-. £ A = 28° 16' 2"
b= 55° 40' log sin (s - a) =^74943^10^ %B = 34° 33' 41.5"
c= 60° 1W log sin (s- 6) =9.64184 -10 I I C = 39° 29' 12"
•. 2 s = 163° 20' log sin (s - c) ^ko-SG^OSj-JB/ .'. A = 56° 32' 4"
81° 40' /. log tan2 r = 18.95996 - 20 B = 69° T 23"
34° 10' /. log tan r = 9.47998 - 10 C = 78° 58' 24"
26° /. log tan | A = 9.73055 - 10
21° 30' log tan \ B = 9.83814 - 10
logtan£C = 9.91590-10
Check : log sin a = 9.86763 log sin b = 9.91686 log sin c = 9.93826
log sin A = Q. 92 128 log sin J5 = 9.97051 log sin C = 9.99191
9.94635 9.94635 9.94635
NOTE 1. Directions for the numerical work: Fill in the first column;
turn up the first four logarithms in the second column (since these logarithms
are required by the formulas) ; compute the last five logarithms in the second
column according to the formulas (these computations may be made on
another paper, if necessary) ; find the first three angles of the third column
by the tables ; thence compute A, B, C.
NOTE 2. If only one angle is required, say A, it can be found by one of
formulas (4) Art. 37 ; preferably, the second. Angle A can also be founcl
(without logarithms) by formula (1) Art. 37,
56 SPHERICAL TRIGONOMETRY. [Cn. IV.
X 2- Solve ABC, given that a = 43° 30', b = 72° 24', c = 87° 50'.
3. Solve ABC, given that a = 110° 40', b = 45C 10', c = 73° SO'.
4. Solve ABC, given that a = 120° 50', 6 = 98° 40', c = 74° 60'.
5. Solve PQR, given that p = 67° 40', q = 47° 20', r = 83° 50'.
43. Case II. Given the three angles.
Either: Solve the polar triangle by the method used in Case L,
and therefrom obtain the parts of the original triangle.
Or: Solve by means of formulas (8)-(ll) Art. 37.
EXAMPLES.
Solve ABC, and check the results.
L^\. Given A = 74° 40', B = 67° 30', C = 49° 50 .
2. Given A = 112° 30', B = 83° 40', C = 70° 10'.
3. Given A = 130°, B = 98°, C = 64°.
4. Given P = 33° 40', $ = 26° 10', R = 20° 30'. Find p, q, r.
NOTE. The results may also be checked by solving the examples by both
the methods above.
44. Case III. Given two sides and their included angle.
EXAMPLES.
1. In ABC, a = 64° 24', b = 42° 30', C = 58° 40' ; find A, B, c.
Formulas : tan $(A + B) = -n Cot \ C ;
cos (a + 6)
sin i O + 6)
sin c = 5*™ Sin a
sin ^4
Checks : Law of Sines, etc.
C = 58° 40' log cot £C=0.25031 .-. log tan ^(A + B)= 0.46743
a= 64° 24' log sin i (a + 6) =9.90490 -10 log tan \(A-B) =9.62405
b = 42° 30' log cos £(« + &) =9.77490- 10 •'• K-4+-B) = 71° 10' 41"
... a-}- 6 =106° 54' logsin £(«- 6) =9.27864-10.. l(A-B)=22°W 12"
a-&=21°54' log cos $(a-6)=9. 99202 -10 /. ^1=93° 59' 53"
|C= 29° 20' logsina=9,95513-10 £=48° 21' 29"
J(a+6)= 53°27' logsin 0=9.93154-10 .-. log sin .4=9. 99894 -1(
J(a-6)= 10° 57 /. log sine =9. 88773 -1(
/. c=60°33'6''
43-46.] THE AMBIGUOUS CASE. 57
NOTE 1. Since C<A, then c<a ; and hence, the acute value of c is taken.
NOTE 2. Directions for the numerical work: Fill in the first column;
then turn up all the logarithms for the second column, these logarithms being
required by the formulas ; then compute the first two logarithms in the third
column, according to the formulas ; thence find the corresponding angles, and
calculate A and B ; turn up log sin A ; compute log sin c according to the
formula ; then find c in the Tables.
NOTE 3. In using formulas involving the difference of two sides or two
angles, place the larger side or angle first.
L 2. Solve ABC, given a = 93° 20', &= 56° 30', C = 74° 40'.
3. Solve ABC, given b = 76° 30', c = 47° 20', ^ = 92° 30'.
4. Solve ABC, given c = 40° 20', a = 100° 30', B = 46° 40'.
5. Solve PQE, given # = 76° 30', r = 110°20', P=46°50'.
45. Case IV. Given one side and its two adjacent angles.
Either : Solve the polar triangle by the method used in Case III. ; and
therefrom obtain the parts of the original triangle.
Or : Solve by using formulas (1), (2), Art. 38.
EXAMPLES.
1. Solve ABC, given A = 67° 30', B = 45° 50', c = 74° 20f .
2. Solve ABC, given I? = 98° 30', C = 61° 20', a = 60° 40'.
3. Solve ABC, given C = 110°, A = 94°, b = 44°.
4. Solve PQE, given E = 70° 20', Q = 43° 50', p = 50° 46'.
46. Case V. Given two sides and the angle opposite one of them.
This is an ambiguous case,* since (Art. 24, V.) there may be two
solutions. It may be well to examine this case (1) geometrically,
that is, by an inspection of the figure ; (2) analytically, that is,
by an inspection of the formulas involved in its solution.
(1) Geometrically. In Art. 24, V. (Figs. 21, 25) it has been
seen that, when two sides and an angle opposite one of them
(say, a, b, A) of a triangle ABC are given, there are two triangles
possible if either of the following sets of conditions holds, viz. :
A < 90°, a >p, a < b, and a < 180° - b ; (a)
A > 90°, a < p, a>b, and a > 180° - b. (b)
* For a detailed discussion of the ambiguous case, see Todhunter, Spher-
ical Trigonometry, pp. 53-58; M'Clelland and Preston, Spherical Trigo-
nometry, pp. 137-143.
58 SPHERICAL TRIGONOMETRY. [Cn. IV.
In order that the triangle be possible, it is apparent that: either
CB=CP-, or, in Fig. 21, CB>CP, i.e. sin CB > sin CP, i.e.
sin a > sin AC sin CAP,
i.e. sin a > sin bsinA-, .
and, in Fig. 25. CZ5 < CP, and CLB > CP1, i.e. sin a>CP', i.e.
sin a > sin AC sin CAP',
i.e. sin a > sin b sin (180° — CAP), i.e. sin a > sm b sin A.
Art. 24 also shows that, when the triangle is possible, there is
one solution if either of the following sets of conditions holds, viz. :
^4<90°, a>p, a between b and 180° -6; (c)
^>90°, a<p, a between b and 180° - b. (d)
If CB = CP, i.e. if a = p, then there is one solution.
Art. 24 also shows that the triangle is impossible if either one
of the following sets of conditions holds, viz. :
A < 90°, a greater than both b and 180° - b j (e)
A > 90°, a less than both b and 180° - b.
Since the greater angle is opposite the greater side, B must be
such that A — B shall have the same sign as a — b.
(2) Analytically. The formulas used in solving this case are as
follows :
sma
7| cot iC=!^|g±| tan l(A - B), [or, (4) Art. 38] (2)
tan*(a~6>- l>. (2) Art 38] (3)
Since B is determined from its sine, it may be in either the first
or the second quadrant. If sin a — sin b sin A, then B = 90°. If
sin a < sin b sin A, then sin B > 1, and B has an impossible value,
and, accordingly, the triangle is impossible. [Compare above. J
Equation (2) shows that A — B and a — b have the same sign.
46.] THE AMBIGUOUS CASE. 59
Hence, from the analytical inspection comes the following rule:
If sin a < sin b sin A9 there is no solution ; if sin a = sin b sin A ,
there is one solution ; if sin a > sin b sin JL, cmcZ if both values of B
obtained from (1) 6e swc/i that
A — B and a — b have like signs,
there are two solutions; if only one o/ the values of B satisfies this
condition, there is only one solution; if neither of the values of B
satisfies this condition, the solution is impossible.
From the geometrical inspection comes the following rule :
If sin a < sin b sin A, there is no solution ; if sin a - sin b sin A,
there is one solution ; if sin a > sin b sin A9 then :
When A is less than 90°:
there are two solutions if a is less than both b and 180° — b;
there is one solution if a lies between b and 180° — b ;
there is no solution if a is greater than both b and 180° — b.
When A is greater than 90°:
there are two solutions if a is greater than both b and 180° — b;
there is one solution if a lies between b and 180° — b;
there is no solution if a is less than both b and 180° — b.
NOTE 1. The second rule has one advantage over the first, in that it
enables one to say, merely on calculating sin B, but without finding J5,
whether the triangle is ambiguous or not.
NOTE 2. The property observed in Art. 36, Note 2, is frequently used in
investigating the ambiguous case.
EXAMPLES.
1. In ABC, a = 43° 20', 6 = 48° 30', A = 58° 40' ; find B, C, c.
Formulas: sin B =.
sin a
C0t * ° = ta" * (B - A) • [Art 38'
tan * " = ri- ta" * (6 - a)' [Art' 38'
Checks: Formulas (2), (4), Art. 38 ; or, formulas, Art. 37 ; or, Delambre's
Analogies.
60 SPHERICAL TRIGONOMETRY. [Cn. IV.
A = 58° 40' log sin A = 9.93154 - 10 .-. B + A = 127° 27'
a = 43° 20' log sin a = 9.83648 - 10 B - A = 10° 7'
6 = 48° 30' log sin b = 9.87446 - 10 %(B + A) = 63° 43' 30"
.-. b + a = 91° 50' .-. log sin B = 9.96952 - 10 %(B-A)= 5° 3' 30"
b-a= 5° 10' .-. B = 68° 47' .-. B' + A = 169° 53'
£(& + «) =45° 55' .B' = lll°13' B'-A= 52° 33'
£(& - a) = 2° 35' [According to the test for $(B' + A) = 84° 56' 30"
ambiguity.] $(B' - A) = 26° 16' 30"
In ABC. (See Fig. 21, Art. 24.) In AB'C.
log sin £(& + a) = 9. 85632 -10 r
log sin £(& - a) = 8.65391 - 10 j As in ABC.
log tan J(6 - a) = 8.65435 - 10
log sin J(B + A) = 9.95264 - 10 log sin %(B' + A) = 9.99830 - 10
log sin J(B - -4) = 8.94532 - 10 log sin $(.B' - A) = 9.64609 - 10
log tan %(B - A)= 8.94702 - 10 log tan J(J3' - ^1) =' 9.69345 - 10
/. log cot J C = 0. 14943 .-. log cot \ C = 0.89586
log tan £ c = 9.66167 - 10 log tan J c = 9.00656 - 10
.-. £C=35°19'55".4, £c=24°38'53". I .-. %C= 7° 14' 36", Jc= 5° 47' 49".
.-. (7= 70° 39' 51", c=49°17'46". I .-. O = 14°29' 12", c=ll°35'38".
Hence, the solutions are :
ABC= 68° 47', ACS = 70° 39' 51", AB = 49° 17' 46" ;
AB' (7=111° 13', ACS' = 14° 29' 12", AB1 = 11° 35' 38".
NOTE 3. Directions for the numerical work : Fill in the first of the three
columns ; turn up the first three logarithms in the second column, these
being required by the first formula ; compute log sin B according to the first
formula ; find B in the tables ; decide the question of ambiguity ; fill in the
third column (only four lines when the triangle is not ambiguous). Turn up
the first six logarithms in the first of the next two columns ; compute the
next two logarithms according to the formulas ; find the corresponding values
\ in the Tables ; thence compute C and c. If the case is ambiguous, do the
V same work for the second triangle.
2. Solve ABC when a = 56° 40', b = 30° 50', A = 103° 40'.
3. Solve ABC when a = 30° 20', b = 46° 30', A = 36° 40'.
4. Solve ABC when c = 74° 20', a = 119° 40', C = 88° 30'.
5. Solve ABC when b = 30° 10', c = 44° 30', B = 86° 50'.
6. Solve PQR when q = 42° 30', r= 46° 50', Q= 56° 30'.
47-48.] SUBSIDIARY ANGLES. 61
47. Case VI. Given two angles and the side opposite one of them.
This is also an ambiguous case.
Either : Solve the polar triangle by the method used in Case V. ;
and therefrom obtain the parts of the original triangle.
Or : Solve by using formula (1) Art. 36, and Napier's Analogies.
The first rule (Art. 46) for determining ambiguity suits the
case, if a, 6, be substituted for A, B, therein. On making use of
the polar triangle, it is found that the second rule can be adapted
by substituting a, A, B, for A, a, b, respectively.
EXAMPLES.
I/I. Solve ABC when A = 108° 40', B = 134° 20', a = 145° 36'.
2. Solve ABC when B = 36° 20', C = 46° 30', b = 42° 12'.
3. Solve ABC when C = 62° 10', A = 23° 46', c = 33° 50'.
4. Solve 8TV when T = 102° 50', F= 81° 20', t = 124° 30'.
48. Subsidiary angles. Formulas can sometimes be adapted for loga-
rithmic computation and the triangle solved, by the use of subsidiary angles.
For example, in ABC let a, c, B be known, and b required. (See Fig. 35,
Art. 33.)
cos b = cos a cos c + sin a sin c cos B (Art. 36, B)
— cos c (cos a -f sin a tan c cos J5).
On putting tan c cos .B = tan 0, this becomes
cos b = cos c (cos a + sin a tan 0)
__ cos c (cos a cos 0 + sin a sin 0)
cos 0
_ cos c cos (q — 0)
cos 0
On referring to Fig. 35 it is seen that BD = 0, that DC = a — 0, and
cos .4Z) = ; so that solving as above is equivalent to solving the triangle
cos 0
by dividing it into right-angled triangles.
N.B. Questions and exercises on Chapter IV. will be found on page 105.
CHAPTER V.
CIRCLES CONNECTED WITH SPHERICAL TRIANGLES.
49. The circumscribing circle. The circle passing through the
vertices of a spherical triangle is called the circumscribing circle,
or drcum-circle, of the triangle. This circle can be constructed
in somewhat the same manner as the circumscribing circle of a
plane triangle.
Let ABC (Fig. 41) be a spherical triangle, and let R denote
the radius (i.e. the polar distance, Art. 6)
of its circumscribing circle. Bisect the
arcs BC, CA, in L, M, respectively; and
at Z, M, draw arcs at right angles to BC,
CA, respectively. The point 0, at which
these arcs meet, is the pole of the circum-
scribing circle.
For, draw OA, OB, OC, arcs of great
circles. In the triangles OLB and OLC,
BL = LC, LO is common, and the angles at L are right angles.
Hence, OB = OC. In a similar way it can be shown that
OC = OA. Hence 0 is the pole of the circumscribing circle.
Join 0 and N, the middle point of AB ; then it is easily shown
that ON is at right angles to AB.
In ABC, A + B+C=2S.
Now (since OA = OB = OC),
OAB=OBA, OBC=OCB, OCA= OAC.
Hence, OAB+ OBC+ OAC=S.
... OBC= S-(OAB + OAC) = S- A.
In the right-angled triangle OBL,
^°B-^L- C^t. 26, E,, (3)]
62
49-50.]
THE INSCRIBED CIRCLE.
63
On substituting in (1) the value of tan J a in relation (8) Art.
37, equation (1) becomes
tan
-V
—cos
cos (# - A) cos (£ - B) cos (S—C)
NOTE 1. Compare (1) with the corresponding case in plane triangles
(Plane Trig., Art. 68). (In plane triangles, £=90°, and, hence,
COB (#--4) = sin 4.)
NOTE 2. On putting N= V- cos /SYcos(# - A)cos(S - B)cos(S - (7),
50. The inscribed circle. The circle which touches each of the
sides of a spherical triangle is called the inscribed circle, or in-
circle, of the triangle. This circle can be constructed in some-
what the same manner as the inscribed circle of a plane triangle.
Let ABC be a spherical triangle, and let
r denote the radius (f..e. the polar distance)
of its inscribed circle. Bisect angles A, B,
by arcs of great circles, and let these arcs
meet at 0. Draw OL, OM, ON, at right
angles to BC, CA, AB, respectively.
In the triangles 0AM and OAN, the
angles at A are equal, the angles at N and
M are right angles, and the side OA is
common. Hence these triangles are symmetrical, and OM '= ON.
Similarly it can be shown that 0^7"= OL. Hence 0 is the pole
of the circle inscribed in ABC.
Since the triangles 0AM and OAN are equal, AM = AN.
Similarly, BN=BL, and CL = CM.
Now AB -}- BC + CA = 2 s •
hence AN + BL + CL = s.
.-. AN= s - (BL +-LC) = s - a.
64 SPHERICAL TRIGONOMETRY. [Cn. V.
In the right-angled triangle AON,
tan ON = tan OAN sin AN. [Art. 26, (4)]
.'. tan r = tan | A sin (s - a) . (1)
Similarly, tan r — tan £ B sin (s — 6) ; tan r = tan 1 (7 sin (s — c).
On substituting in (1) the value of tsm^A in (4) Art. 37,
equation (1) becomes
tan r
= A/ sin O •—
*
sin *
On putting n = Vsin s sin (s - a) sin (« - ft) sin (« - c) , (3)
tanr = -^-. (4)
sin s
NOTE 1. Compare (1) with Plane Trigonometry, Art. 69, Note ; (2) with
Art. 69, (3); n with S, Art. 66, (3); (4) with (3) Art. 69.
51. Escribed circles. A circle which touches a side of a spher-
ical triangle, and the other two sides produced (that is, which
is inscribed in a co-lunar triangle), is an escribed circle, or an ex-
circle, of the triangle. There are three ex-circles, one correspond-
ing to each side of the triangle.
Let ABC be a spherical tri-
angle ; and let the radii of the
escribed circles, touching a, b, c,
respectively, be denoted by ra, rb,
rc, respectively. Complete the
lune whose angle is A. The
escribed circle which touches a is the inscribed circle of the
co-lunar triangle ABC. Hence [Art. 50, (1)],
tan ra = tan £ A' sin } [(a + 180° - b + 180° - c) - 2 a] ;
i.e. tan ra = tan | A sin s. (1)
Similarly, tan r6 = tan ± B sin s ; tan rc = tan £ C sin s.
51.] ESCRIBED CIRCLES. 65
On substituting for tanjyl its value in (4) Art. 37, equation
(1) becomes
tan r =
sin (s — a)
tann- = [^t. 50, (3)] (3)
Similarly, tan rb = — — -^ ; tan rc = —
sin (s — 6) sin (s — c)
NOTE. Compare (3) with the corresponding result in Plane Trigonome-
try, Art. 70.
Some other relations between the sides and angles of a spherical
triangle and the radii of the circles connected with it} are indicated
in the exercises at the end of the book.
Ex. Find the radii of the circumscribing, inscribed, and escribed circles of
some of the triangles in Chapters II., IV.
N.B. For questions and exercises on Chapter F., see page 107.
CHAPTER VI.
AREAS AND VOLUMES CONNECTED WITH SPHERES.
52. Preliminary propositions.
a. The lateral area of a frustum of a regular pyramid is equal
to the product of the slant height of the frustum and half the
sum of the perimeters of its bases.
Ps
re,,
LK-'-iO
J T>
Fia. 45
FlO.
The student can easily prove this (Fig. 44). It should be
noted that the half sum of the perimeters of the bases of the
frustum is equal to the perimeter of the section which is parallel
to the bases and midway between them.
In symbols: If p^ p.2, P, are the perimeters of the. bases and
the middle section of the frustum, and MN is its slant height,
theii
lateral area of frustum = 1 MN (pl + p>l) = MN - P.
6. The lateral area of a frustum of a cone of revolution is
equal to the product of the slant height of the frustum and half
the sum of the circumferences of its bases.
[Suggestion for proof: If the number of the lateral faces of a
frustum of a regular pyramid be indefinitely increased and each
face be indefinitely decreased, then this frustum approaches the
frustum of a cone of revolution as a limit (see Fig. 46). Accord-
ingly, Proposition (b) follows at once from (a)]. It should be
52-53.] AREA OF A SPHERE. 67
noted that half the sum of the circumferences of the bases of.
the frustum is equal to the circumference of the section which
is parallel to the bases and midway between them.
In symbols : If Q, C2, C (Fig. 45) are the circumferences of the
bases and the middle section of the frustum, and MN is its slant
height, then lateral area of frustum
= | MN (Ci + Ci) = MN- C=2 TrLG - MN.
NOTE: The lateral surface of the frustum of the cone (Fig. 45) can be
generated by the revolution of the line MN about the line AB which is in
the same plane with MN.
53. To find the area of a sphere. The surface of a sphere can
be generated by the revolution of a semicircle about its diameter.
For example, the semicircle ATKB of radius R
on revolving about its diameter AB, will describe
the surface of a sphere of radius OA.
Let a polygon ALTGKB be inscribed in this
semicircle. At M, the middle point of one of the
chords LT, draw MO at right angles to LT. By Q
geometry, MO will meet AB at 0, the middle
point of AB. Project LT on AB, the- projection
being It ; draw LQ at right angles to Tt.
By Art. 52. 6, the area generated by LT in its revolution
about AB
= 2 irMm.LT. (1)
Since the angles of the triangle LTQ are respectively equal to
the angles of OMm, these triangles are similar ; accordingly,
LT:LQ= OM:Mm.
.-. Mm-LT=LQ. OM = It . OM.
Hence, from (1), area generated by LT = 2 vOM • It. (2)
In words: When a chord of a semicircle revolves about the
diameter, the area generated is equal to 2 TT times the product of
the length of the perpendicular from the centre to the chord,
and the projection of the chord upon the diameter.
68 SPHERICAL TRIGONOMETRY. [Cn. VI.
.*. The area of the surface generated by the revolution of the
polygon ALTGKB
= 2 TT x (perpendicular on AL from 0) X Al
-|- 2 TT x (perpendicular on LT from 0) x It
+ 2 TT x (perpendicular on TG from 0) x tg
4- 2 TT x (perpendicular on GKfrom 0) x #&
+ 2 TT x (perpendicular on KB from 0) x kB.
If the number of sides in the polygon inscribed in the semi-
circle is indefinitely increased and each side is indefinitely de-
creased, then the broken line ALTGKB approaches the semicircle
as a limit, and each of the perpendiculars drawn from 0 to the
middle points of the chords approaches R as a limit ; while
the sum of the projections of the chords remains equal to AB,
the diameter of the circle. Hence, area of surface generated by
revolution of semicircle AGB = 2 -*• • R • 2 R ;
i.e. area of surface of sphere of radius R = 4 wR2.
In words : The area of the surface of a sphere is four times the
area of a great circle of the sphere.
Definition. A zone of a sphere is a portion of the surface in-
cluded between two parallel planes, or, what comes to the same
thing, is the portion of the surface included between two circles
which have common poles ; for example, the surface between the
parallels of 30° N. latitude and 50° K latitude.
The area of a zone. An infinite number of chords can be in-
scribed in the arc LT (Fig. 47). By reasoning similar to that
employed above, it can be shown that
area of surface generated by arc LT= 2 irR • It.
.-. The area of a spherical zone is equal to the product of the length
of a great circle of the sphere and the height of the zone.
It follows that on a sphere or on equal spheres the areas of
zones of equal heights are equal.
54-55.] A SPHERICAL DEGREE DEFINED. 69
EXAMPLES.
1. Find the area of a sphere of radius 15 inches.
2. Find the surface of a spherical zone of height 2.5 inches on a sphere of
diameter 50 inches.
3. Find the convex surface of a spherical segment of height 4.5 inches on
a sphere of diameter 7 feet. [See definition, Art. 63.]
4. Suppose that the earth is a sphere whose radius is 3960 miles ; find the
area of the surface included between the North Pole and the parallel of 80°
N. latitude ; between the parallels of 49° N. and 50° N. ; between 6° N.
and 5° S.
54. Lunes. Definition. The spherical surface bounded by two
halves of great-circles is called a lune; e.g. the surface between
two meridians. The angle of the lune is the angle between the
two semicircles ; thus the angle of the
lune between the meridians 70° W. and
80° W. is 10°.
Proposition. On the same circle or on
equal circles the areas of lunes are propor-
tional to their angles. This can be proved
by a method similar to that which is used
in proving that the angles at the centre of
a circle are proportional to the arcs sub- FIG. 48
tended by them.
55. A spherical degree defined. From the proposition in Art. 54
it follows that the area of a lune is to the area of the surface
of the sphere as the angle of the lune is to four right angles.
That is,
area of lune of angle A° : area of sphere = A° : 360°.
Hence, area of lune of angle 1° = area of sPhere.
360
Let a great circle be drawn about one of the vertices of a lune
of angle 1° as a pole. The lune is then divided into two equal bi-
rectangular triangles ; accordingly, each triangle contains (7J^)th
of the surface of the sphere, or (^^th of the surface of the
hemisphere. The surface of each such triangle is called a spherical
degree.
70 SPHEEICAL TRIGONOMETRY. [Cn. VI.
For example, the part of the surface of a globe bounded by the
meridians 43° W. and 63° W. longitude and the equator, contains
20 spherical degrees ; the lune bounded by these meridians con-
tains 40 spherical degrees.
A lune of angle A° contains 2 A spherical degrees.
The passage from spherical degrees of surface to the ordinary
measure (of the area) of the surface is easily effected when the
radius of the sphere is given.
A spherical degree = (^^)tli part of the surface of a sphere;
hence, on a sphere of radius r,
a spherical degree contains ^TT' i-e- - •— square units of area.
Thus,
area of a lune of angle 20° on a sphere of radius r = — — = f Trr2.
180
EXAMPLES.
1. Find the area of a lune of angle 10° on a sphere of radius 2 feet.
2. Find the area of a lune of angle 37° 30' on a sphere of radius 7 feet.
3. Find the area between the meridians 77° W. and 83° 20' W. ; and the
area between the meridians 174° 20' W. and 158° 35' E. (Radius of earth
= 3960 miles.) [Express areas in spherical degrees and in square miles.]
56. Spherical excess of a triangle. The sum of the angles of a
plane triangle is always equal to 180° ; the sum of the angles of
a spherical triangle is always greater than 180° (Art. 17). The
difference between the latter sum and 180° is called the spherical
excess of the triangle. (This excess is due to the fact that the
triangle is spherical and not plane; hence the excess is called
spherical.) For example, in the triangle bounded by the meridi-
ans 47° W. and 48° W. longitude and the equator, the sum of the
angles is 181° ; and, accordingly, the spherical excess is 1°. In
the triangle bounded by the meridians 43° W. and 63° W. and
the equator the sum of the angles is 200°, and the spherical ex-
cess is 20° ; in the spherical triangle having angles 50°, 65°, 125°,
the spherical excess is (50° + 65° + 125° - 180°), i.e. 60°.
56-57.]
AREA OF A SPHERICAL TRIANGLE.
71
If E denote the number of degrees in the spherical excess, and
Er denote the number of radians therein, then
in a triangle ABC, E° = A° + B° + C° - 180° ;
and [Plane Trigonometry, Art. 73, (7)],
180
(1)
(2)
Ex. Find the spherical excess (in degrees and in radians') of the tri-
angles described in Art. 42, Exs. 1, 2, 3 ; Art. 43, Exs. 1, 2 ; Art. 44, Exs.
1, 2, 3 ; Art. 45, Exs. 1, 2; Art. 46, Exs. 1, 2, 3 ; Art. 47, Exs. 1, 2.
57. The area of a spherical triangle.
Proposition : The number of spherical degrees (of surface) in a
spherical triangle is equal to the number of (angular) degrees in
its spherical excess*
Let ABC be a spherical triangle whose
spherical excess is E° ; then area ABC
is equal to E spherical degrees. Com-
plete the great circle BCB'C', and pro- /
duce the arcs BA, CA to meet this circle
in B', C', respectively. Complete the
great circles BAB'B and AC AC'. The
triangle AB'C' is equal to the triangle
ABC. For,
Fia. 49
'A = 180°-AC=CA,
= ISO°-B'C=CB.
Hence, in area, ABC + AB'C ' = lime ACA'BA;
also ABC + AB'C = lune BCB'AB-,
and ABC + ABC' = lune CBC'AC.
* This proposition is sometimes stated thus : The area of a triangle is
equal to its spherical excess • but this enunciation is rather slipshod.
72 SPHERICAL TRIGONOMETRY. [On. VI.
Hence, on addition,
2 ABC + (ABC + AB'C' + AB'C + ^LBC")
= lime ^4 -f- lune J5 + lune (7;
2 ABC= lune ^4 + lune jB + lune C — hemisphere.
.-. (by Art. 55) 2 ABC = (2 A+2 B+2 C-360) spherical degrees.
ABC = (A + 5 + (7 - 180) spherical degrees
=== .JEJ spherical degrees.
Since (Art. 55) a spherical degree on a sphere of radius r con-
tains T!"jy ?rr2 square units of area, then, on this sphere,
area ABC = A + B^"19° -r»* = j^ -r*> C1)
= J?rr2> [Art. 56 (2)] (2)
in which 1£ denotes the number of degrees, and Er denotes the
number of radians in the spherical excess.
Hence, in order to find the area of a triangle, find the angles,
calculate the spherical excess in degrees or radians, and use one
of formulas (1), (2).
NOTE. It should be observed that [from Art. 14, Art. 56 (1), and the
proposition above], the number of spherical degrees contained in the area
subtended on a spherical surface by a solid angle at the centre of the sphere,
remains the same, however the radius may vary. On the other hand, by (1)
and (2), the number of square units in the subtended area varies as the
square of the radius.
* This expression for the area of a spherical triangle was first given in
1629 by Albert Girard (1590-1634) (see Plane Trigonometry, pp. 22, 167);
and it is often called GirarcTs Theorem. The method of proof used above
was invented by John Wallis (1616-1703) professor of geometry at Oxford.
(See Wallis, Work?, Vol. II., p. 875.)
It follows from (1) that
area ABC : 2 irR* = E° : 360°.
Hence, the above proposition may be expressed thus : The area of a
spherical triangle is to the surface of the hemisphere as the excess of its three
angles above two right angles is to four right angles.
58.] SPHERICAL EXCESS OF A TRIANGLE. 73
EXAMPLES.
Find the areas of the following triangles (see examples, Art. 56) :
1. Those described in Art. 42, Exs. 1, 2, 3, when on a sphere of radius
10 feet.
2. Those described in Art. 43, Exs. 1, 2, when on a sphere of radius
25 inches.
3. Those described in Art. 44, Exs. 1, 2, 3, when on a sphere of radius
30 yards.
4. Those described in Art. 45, Exs. 1, 2, when on a sphere of radius
4 feet.
5. Those described in Art. 46, Exs. 1, 2, 3, when on a sphere of radius
18 inches.
6. Those described in Art. 47, Exs. 1, 2, when on a sphere of radius
3960 miles.
58. Formulas for the spherical excess (£°) of a triangle. Since, in
a spherical triangle ABC, E° = A° + B° -f C° - 180°, and since there are
many relations between the sides and angles of a triangle, it may be expected
that there can be many formulas for the spherical excess ; and, accordingly,
for the area of a spherical triangle. [It will be remembered that there
are several formulas for the area of a plane triangle (Plane Trigonometry,
Art. 66).] Following are some of the most important of these (the deduc-
tion of some of them is given in Note JB) :
A. The spherical excess in terms of the three sides.
(a) L^Huillier's formula :
tan £ E° - Vtan \ s tan |(s - a) tan £(s - 6) tan |(s - c).
(6) CagnolVs formula : sin \ E° =
n
2 cos \ a cos i b cos \ c
in which . n = vsinssin (s — a)sin(s — 6)sin(s — c).
(c) De Gun's formula*: cot ^ = l + COS(* + COB b + COSC.f
* Simon L'lluillier (1750-1810), a Swiss mathematician and philosopher ;
Antoine Cagnoli (1743-1816), an Italian astronomer; L'abbe Jean Paul
de Gua (1712-1786), a French philosopher.
t For the deduction of this formula see Chauvenet, Trigonometry, p. 230,
and Crawley, Trigonometry, p. 166.
74 SPHERICAL TRIGONOMETRY. [On. VI.
B. The spherical excess in terms of two sides and their included angle.
(d) tan i E° = _J^Li^tan_^sin(7
1 + tan I a tan £ b cos O
(e) cot i ^° - cot ^ a cot ^-f cos (7
sin O
Ex. By these formulas find the spherical excess of some of the triangles
referred to in Ex. 1, Art. 56.
59. a. The number of spherical degrees in any figure on a sphere,
whatever may be its boundary, is the ratio of the area of the
figure to the area of a spherical degree, that is, to (^J^)th part of
the area of the hemisphere (Art. 55). Thus, on a sphere of radius
r, if A denotes the area of the figure, and E the number of spheri-
cal degrees therein, then, since area of a hemisphere =
[Compare Art. 57 (1), Art. 59 (2).]
The plane angle E° may be called the spherical excess of the
figure. For example, the spherical excess of a lune of angle A°
is 2A°.
b. The spherical excess of a (non-re-entrant) spherical polygon.
On drawing diagonals from any vertex of a polygon of n sides to
the other vertices, it will be seen that the polygon is divided into
n — 2 triangles. The sum of the angles of all these triangles is the
same as the sum of the angles of the polygon. Hence,
spherical excess (E°) of polygon ofn sides
= sum of angles -(n-2) ISO0.
If the radius of the sphere is r, then (Art. 57)
W
area of the polygon = -— Trr2. (2)
J_Ow
60. Given the arsa of a figure : to find its spherical excess. More
fully : To find the spherical excess of a figure on a sphere when the
area of the figure is given in square units.
59-60.] SPHERICAL EXCESS OF FIGURES. 75
Let r denote the radius of the sphere, A the area of the figure,
E the number of degrees, n the number of seconds, and Er the
number of radians, in its spherical excess. Then, by (1) Art. 59,
.-. n = 3600 E = 206265^,- (2)
Now 1° = -^- radians ;
loU
hence E° = -~:E radians
180
= ^ radians. [by (1)]
•'• E'=f (3)
A particular application of (2) can be made to the following
problem, viz. : The area of a spherical triangle on the earth's sur-
face being known, to derive a formula for computing the spherical
excess.
The length of a degree on the earth's surface is found to be
365155 feet. Accordingly,
R (the radius of the earth) = 365155 x 18° feet. (4)
Prom (2), log n = log A + log 206265 - 2 log R. (5)
On expressing A in square feet, and substituting in (5) the
value of R in (4), there is obtained,
log n = log A - 9.3267737. (6)
Formula (6) is called Roy's Rule, as it was used by General
William Roy (1726-1790) in the Trigonometrical Survey of the
British Isles.* The area of the spherical triangle can be approxi-
mately determined to a sufficient degree of accuracy.
*The rule should probably be credited to Isaac Dalby (1744-1824), who
was mathematical assistant to General Roy from 1787 to 1790, and later
became professor of mathematics at the Royal Military College. [See Phil.
Trans., vol. 80 (1790).] This was the first practical application of Gerard's
theorem (Art. 57).
76 SPHERICAL TRIGONOMETRY. [Cn. VI.
61. The measure of a solid angle. A plane angle can be measured
by any circular arc which it subtends; and the measure can be
expressed in radians and in degrees. The radian (or circular)
measure of an angle is the number of times any circular arc sub-
tended by it contains the radius (Plane Trig., Art. 73) ; and the
number of degrees in the angle is equal to the number of degrees in
the subtended circular arc. Thus, the radian measure of an angle
of an equiangular triangle is ITT, and its degree measure is 60.
A solid angle can be measured in a somewhat similar manner,
namely, by means of any spherical surface which it subtends.
What may be called the spherical measure of a solid angle is the
number of times any spherical sur-
face subtended by it contains an
area equal to the square on the radius.
For example, since the surface of a
sphere is equal to 4^, the sum of
all the solid angles about any point
is 47T. The angle at the corner of
a cube subtends one-eighth of the
FIQ. 50 surface of the sphere ; accordingly, its
spherical measure is -^- -f- r2, i.e. ^ TT.
A solid angle may also be measured in spherical degrees, a term
that will be explained presently. What may be called the
spherical degree measure of a solid angle (or, the number of
spherical degrees in the angle) is a number equal to the number
of spherical degrees of area in any spherical surface subtended
by the angle. An angle that subtends a spherical degree of
surface, contains what may be called a solid spherical degree.
For example, the sum of all the solid angles about any point
is 720 spherical degrees (of angle); the angle at the corner of
a cube contains 90 spherical degrees (of angle). Thus the
spherical measure of the angle at the corner of a cube is ^ TT, and
its spherical degree measure is 90. On comparing these defini-
tions of solid angular measures with Art. 55 and equations (3) and
(1) Art. 60, it is seen that these measures of solid angles are equal
to the measures, in radians and degrees respectively, of the spherical
excess of the Jigures subtended on any sphere by the angle, when
the vertex of the angle is at the centre of the sphere.
61.] MEASUREMENT OF SOLID ANGLES. 77
NOTE 1. The term degree. In geometry and trigonometry the word
degree is used in connection with four very different kinds of quantities ;
namely, circular arcs, plane angles, spherical surfaces, and solid angles.
A degree of arc, or an arcual degree, is (j^)th part of any circle ;
A degree of angle, or an angular degree, is (^|^)th part of four right
angles ;
A degree of surface on a sphere, or a spherical degree of surface, is (7^)th
part of the surface of any sphere ;
A degree of solid angle, or a solid spherical degree, is (7^)th part of the
solid angles about any point.
NOTE 2. If two plane angles are equal, they can be superposed, the one
on the other. On the other hand, just as two figures on a sphere may be
equal in area and differ in every other respect, so two solid angles can be
equal in measure and differ in every other respect.
NOTE 3. The following remarks relating to the measurement of solid
angles are from Hutton's Course in Mathematics, Vol. II., p. 64 :
" Solid angles : If about the angular point of a solid angle as centre, a
sphere be described to radius unity, the portion of its surface intercepted
between the planes which contain the solid angle is the measure of the
solid angle. (This method of estimating the magnitude of solid angles
appears to have been first given by Albert Girard in his Invention Nouvelle
en Algebre, 1629 ; and it would very naturally suggest itself as one of the
simplest applications of his theorem for the spherical excess.)" [Compare
Plane Trigonometry, p. 126, Note 2.]
Ex. 1. The edge angles of a triedral angle are 74° 40', 67° 30', 49° 50' ;
calculate its spherical degree measure, and its spherical measure. (See Ex.
1, Art. 43.)
Ex. 2. The face angles of a triedral angle are 47° 30', 55° 40', 60° 10' ;
calculate its spherical degree measure, and its spherical measure. (See Ex.
1, Art. 42.)
Ex. 3. Two face angles of a triedral angle are 64° 24 , 42° 30', and the
edge angle between their planes is 58° 40' ; calculate its spherical degree
measure, and its spherical measure. (See Ex. 1, Art. 44.)
Ex. 4. A face angle of a triedral angle is 74° 20', and the two adjacent
edge angles are 67° 30' and 45° 50' j calculate its measure. (See Ex. 1, Art.
45.)
Ex. 5. Calculate the spherical degree measure, and the spherical measure,
of the solid angles corresponding to the spherical triangles described in
Art. 42, Exs. 2, 3 ; Art. 43, Ex. 2 ; Art. 44. Exs. 2, 3 ; Art. 45, Ex. 2 j Art.
46, Exs. 2, 3 ; Art. 47, Ex. 2. (See Ex., Art. 56.)
78 SPHERICAL TRIGONOMETRY. [Cn. VI.
62. The volume of a sphere. In some works on solid geometry
and in books on mensuration it is shown that the volume of a
pyramid is equal to one third the product .-:J its base and altitude.
Now suppose that a polyedron (i.e. a solid bounded by plane faces)
is circumscribed about a sphere, each of the faces of the polye-
dron, accordingly, touching the sphere. This polyedron may be
regarded as made up of pyramids which have a common vertex
(namely, the centre of the sphere), and a common altitude (namely,
the radius of the sphere), and which have the faces of the poly-
edron as bases. Then, R being the radius of the sphere,
Vol. of polyedron = ^ R x (sum of faces of polyedron). (1)
If the number of faces of the polyedron be increased and the
area of each face be decreased, then the sum of the faces becomes
more nearly equal to the area of the surface of the sphere, and
the volume of the polyedron becomes more nearly equal to the
volume of the sphere. By increasing the number of faces and
decreasing the area of each face, the difference between the sum
of the faces of the polyedron and the area of the sphere can be
made as small as one please ; and, likewise, the difference between
the volume of the polyedron and the volume of the sphere can be
made as small as one please. In other words :
The area of the surface of the sphere is the limit of the area of
the surface of the polyedron, and the volume of the sphere is the
limit of the volume of the polyedron, when the faces of the latter
are increased without limit, and each face is made to approach
zero in area.
Hence, from (1), Vol. of sphere = \ R x surface of sphere
(2)
63. Definitions. A spherical pyramid is a portion of a sphere
bounded by a spherical polygon and the planes of the sides of the
polygon. The polygon is called the base of the pyramid.
* For a note concerning the measurement of the circle and the sphere see
Plane Trigonometry, Art. 72, and Note C, p. 171. For the proofs of Archi-
medes, see T. L. Heath, The Works of Archimedes edited in modern notation,
with introductory chapters (Cambridge, University Press), pp. 39, 41, 93.
62-64.] VOLUME OF A SPHERICAL PYRAMID. 79
For example, in Fig. 11, Art. 12, 0-ABCD, 0-ABC, 0-ABD,
are spherical pyramids ; their bases are ABCD, ABC, ABD.
A spherical sector is the portion of a sphere generated by the
revolution of a sector of a circle about any diameter of the circle
as axis. For example, in Fig. 47, Art. 53, when the semicircle
ATB revolves about AB, each of the circular sectors AOL, LOT,
LOK, etc., describes a spherical sector.
A spherical segment is the portion of a sphere bounded by two
parallel planes and the zone intercepted between them. (One of
the planes may be tangent to the sphere.)
64. Volume of a spherical pyramid ; of a spherical sector. By
reasoning analogous to that in Art. 62, it can be shown that, in a
sphere of radius R,
vol. of a spherical pyramid = ^ R x area of its base ;
vol. of a spherical sector = J R x area of its zone.
Since the area of a zone of height h = 2 -rrRh (Art. 53),
then vol. of spherical sector = f irl&h.
Thus in Fig. 11, Art. 12,
vol. 0-ABCD = i OA x area ABCD;
in Fig. 47, Art. 53,
vol. of sector described by AOL = ± OA x area of zone described
by arc AL = f TT R2 - Al, and
rol. of sector described by LOT = 1 OA x area of zone described
by arc LT = £ irR2 - It.
EXAMPLES.
1. Find the volumes of the spherical pyramids whose bases are the tri-
angles described in Art. 57, Exs. 1-6.
2. Find the volumes of the following spherical sectors :
(a) The sector whose base is a zone of height 2 inches on a sphere of
radius 18 inches.
(6) The sector whose base is a zone of height 3 feet on a sphere of radius
12 feet.
80
SPHERICAL TRIGONOMETRY.
[Cn. VI.
65. Volume of a spherical segment. Let AB be an arc of a semi-
circle of radius R having the diameter DD'. From A, B, draw
Aa, Bb, at right angles to DD'. It is required to find the volume
of the spherical segment generated by the revo-
lution of ABba about DD'.
Let h denote the height of the segment, and
PD P2> the lengths of the perpendiculars from
the centre 0 to the parallel bases of the seg-
ment. On making the revolution of the semi-
circle DAD', it is seen that
segment generated by ABba = cone generated
by BOb + spherical sector generated
by AOB — cone generated by A Oa.
Now, vol. cone generated by BOb = 1 7rr22p2;
vol. sector generated by AOB = f irR^h ; (Art. 64)
vol. cone generated by AOa = ^-n-Tip^
.-. vol segment = 1 TT (r£p2 -f 2 R~h - rfa). (1)
NOTE. The result (1) can be reduced to various forms. For example,
since
then vol. segment = §*• R*(pt -j>0+ | *ps(R* -p22)-
Since
h=p2— p\, then 7i2 = p22 — 2 j92 px -f pi2.
+ P-V, and
On substituting the last result in (3), expressing p^ and p£ in terms of
, ri, r2, and reducing, the following formula is obtained, viz. :
vol. segment =
+ r22 +
(4)
65.] VOLUME OF A SPHERICAL SEGMENT. 81
EXAMPLES.
1. Show that if (in Fig. 51) angle AOD = «, then the volume of the
spherical sector generated by AOD is f TT R»(\ — cos a).
2. Show that if angle AOD = «, then the volume of the segment generated
by the revolution of ADa is | wR3 sin* \ a(l + 2 cos2 1 a).
SUGGESTION. Segment generated by ADa = sector generated by AOD —
cone generated by AOa.
3. Find the volume of a spherical segment, the diameters of its ends being
10 and 12 inches, and its height 2 inches.
4. The diameters of the ends of a spherical segment are 8 and 12 inches,
and its height is 10 inches. Find its volume.
N.B. For questions and exercises on Chapter VI. , seepage 108.
CHAPTER VII.
PRACTICAL APPLICATIONS.
66. Geographical problem. To find the distance between two places
and the bearing (i.e. the direction) of each from the other, when their
latitudes and longitudes are known. An interesting application of
spherical trigonometry can be made in solving this problem. In
the following examples the earth, is regarded as spherical, and its
radius is taken to be 3960 miles.
EXAMPLES.
1. Find the shortest distance along the earth's surface hetween Baltimore
(lat. 39° 17' N., long. 76° 37' W.) and Cape Town (lat. 33° 56' S., long.
18°26'E.).
In Fig. 52 B and C represent Baltimore and Cape Town ; EQ is the earth's
equator ; NGS, NBS, NCS are the meridians of
Greenwich, Baltimore, and Cape Town respec-
tively ; BC is the great circle arc whose length is
required.
In the spherical triangle BNC, NB, JVC, and
BNC are known. For
NB = 90° - BL = 90° - 39° 17' = 50° 43'
NO = 90° + TO = 90° + 33° 56' = 123° 56'
BNC = BNG + GNC = 76° 37' + 18° 26' = 95° 3'
Hence, BC can be determined in degrees by Art. 44 ; then, the radius
of the sphere being given, BC can be determined in miles. The angles NBC,
NCB, can also be found.
Answers : BC = (65° 47' 48") = 4685.8 miles ; NBC = 115° 1' 35' ; NCB
= 57° 42' 23".
NOTE 1. The bearing of one place from a second place is the angle which
the great circle arc joining the two places makes with the meridian of the
second place. Thus, in Fig. 52 the bearing of Cape Town from Baltimore is
the angle NBC, and the bearing of Baltimore from Cape Town is NCB.
82
66.] PRACTICAL APPLICATIONS. 83
Since NBC = 115° lf 35" the ship sets out from Baltimore on a course
S. 64^ 58'' 25" E.; since NCB=bl° 42' 23" the ship approaches Cape Town
on a! course S. 57° 42' 23" E.
NOTE 2. A ship that sails on a great circle (excepting the equator or a
meridian) must be continually changing her course.
2. Find the latitude of the place where BC crosses the meridian 15° W. ;
also find the bearing of Cape Town from this place.
3. If a vessel sails from Baltimore and keeps constantly on the course
(see Ex. 1) S. 64° 58' 25" E. (i.e. crosses every meridian at the angle 64° 58'
25"), will she arrive at Cape Town? [Answer. No.]
4. What path will the vessel in Ex. 3 make on the sea ? Answer. A
path which is a spiral going round and round the earth and gradually
approaching the south pole. This path is called the loxodrome, or rhumb
line.
5. If a person leaves Boston, Mass. (lat. 42° 21' N., long. 71° 4' W.), start-
ing due east, and keeps on a great circle : (a) Where will he be after he has
passed over an arc of 90°, and in what direction will he be going ? (&) Where
will he be after he has passed over an arc of 180°, and in what direction will
he be going ? (c) Where will he be after he has passed over an arc of 270°,
and in what direction will he be going ? [Solve this example : (1) by
spherical geometry ; (2) by spherical trigonometry.]
6. What is the distance from New York (40° 43' N. , 74° 0' W. ) to Liverpool
(53° 24' N., 3° 4' W.)? Find the bearing of each place from the other. In
what latitude will a steamer sailing on a great circle from New York to Liver-
)1 cross the meridian of 50° W. ; and what will be her course at that point?
N.B. Check the results in the following exercises :
7. Find the distance and bearing of Liverpool from Montreal (45° 30' N.,
73° 33' W.).
8. Find the distance and bearing of Liverpool from Halifax, N. S. (44°
40' N., 63° 35' W.).
9. Find the distance and bearing of Santiago de Cuba (20° N., 75° 50' W.)
from Rio de Janeiro (22° 54' S., 43° 8' W.).
10. Find the distance and bearing of San Francisco (37° 47' 55" N.,
122° 24' 32" W.) from New York.
11. Find the distance of Victoria, B. a (48° 25' N., 123° 23' W.) from
Sydney, N. S. W. (33° 52' S., 151° 13' E.) ; and the bearing of each place
from the other.
84 SPHERICAL TRIGONOMETRY. [Cn. VII.
12. Find the distances between the following places : (a) San Francisco
and Honolulu ; (&) Cape Town and Cairo ; (c) Honolulu and Manila ;
(d) Victoria, B. C., and Tokio.
13. Find the distances between other places, and their bearings from each
other.
^
APPLICATIONS TO ASTRONOMY.
N.B. In connection with his study of the following articles the
student should consult some elementary text-book on astronomy. The
numerical examples given here will supplement his outside reading on
spherical astronomy.
67. One of the most important applications of spherical trigo-
nometry is to astronomy. Trigonometry was invented to aid
astronomy, and for centuries was studied as an adjunct of the
latter subject. (See Plane Trigonometry, pp. 165, 166.) A few of
the simplest problems of spherical astronomy are introduced in
Arts. 73, 74. In order to understand these problems a clear con-
ception of a few astronomical terms and principles is necessary.
These terms are explained in Arts. 68-72.
68. The celestial sphere. To a person on the surface of the
earth, the sky above is like a great hemispherical bowl with him-
self at the centre. The stars seem to move from east to west
across the spherical sky in parallel circles whose axis is the earth's
polar axis prolonged. Each star makes a complete revolution
about this axis in 23 hours 56 minutes ordinary clock time. The
stars appear never to change their positions with reference to one
another, being in this respect like places on the earth's surface.*
Another way of describing the relations of the earth and the
enveloping sky, is to say that the whole sky is turning, like an
immense crystal sphere, about an axis which is the earth's polar
axis prolonged, the motion being from east to west. The stars
keep the same positions with respect to one another, and, accord-
ingly, appear to be attached to the surface of the sphere. As the
sphere turns, the stars fixed in it appear to trace parallel circles
* The positions of some of the stars suffer a very slight change which is per-
ceptible in the course of centuries.
67-68.] APPLICATIONS TO ASTRONOMY. 85
about the axis. The sphere turns completely in 23 hours 56
minutes ordinary clock time.* The stars all seem to be at the
same distance from the observer because his eyes can judge their
directions only, and not their distances.
The following considerations will show that it is natural enough for an
observer on the earth to think that he is always at the centre of the sphere on
which the stars appear to be. When a person changes his position, the direc-
tion of an object at which he is looking changes also, unless he moves directly
towards or away from the object. For instance, from a certain point a tree
may be in an easterly direction, and when the observer moves a little way
the tree may be in a southeasterly direction. Moreover, the further away an
object is, the less will be the change in its direction caused by any particular
change in the observer's position. Thus, if a person is near a tree, a few steps
on his part may change the direction of the tree from east to southeast, but if he
is five miles from the tree, an equal number of steps taken by him will make
very little difference in the direction of the tree. Now the earth's mean distance
from the sun is about 93,000,000 miles. Hence, an observer who now looks at
the stars from a certain position, in about six months from now will look at
them from a point 186,000,000 miles distant from his present position, t As-
tronomers have succeeded in a few instances in determining the distances of
the stars from the earth, t It has been found that the nearest star yet known,
Alpha Centauri, is so far away that the change in its direction from the centre
df the earth, due to the change of position of 186,000,000 miles on the part of
the earth, is less than the change in the direction of an object 3£ miles away
when the observer moves his head a couple of inches at right angles to the
line of sight. This being so in the case of the sun's nearest stellar neighbour, it
is natural for an observer on the earth to think that he is always at the centre
of the great sphere on which the stars appear to be ; and it is perfectly proper
* The student probably knows that the apparent turning of the spherical
sky from east to west about an axis which is the earth's polar axis prolonged,
is really due to the rotation of the earth in an opposite direction. The
observer is not conscious of any motion of the, earth, and thinks that the sky
with its bright points is revolving about the earth from east to west, while all
the time the sky is motionless, and the earth is turning under it from west to
east. Just as to a person in a swiftly moving train the objects outside seem
to be rushing by him while the train appears to be at rest.
t This, moreover, does not take any account of the motion of the sun with
his system through space.
J The first stellar distance determined was that of 61 Cygni by Friedrich
Wilhelm Bessel (1784-1846), one of the greatest of German astronomers, in
1838. Si-nce then the distances of about 100 stars have been measured ; about
50 of these distances are regarded as reliably determined.
86 SPHERICAL TRIGONOMETRY [Cn. VII.
for him to act in accordance with this notion when he makes astronomical
observations and deductions.*
Thejsphere on which the stars appear to move in parallel cir-
cles, or, what comes to the same thing, the sphere which appears
to have the stars attached to it and to revolve about the earth's
polar axis prolonged, is called the celestial sphere.
69. Points and lines of reference on the celestial sphere. There
will now be shown some methods for indicating the positions of
the heavenly bodies on the celestial sphere — their positions with
respect to the observer and their positions with respect to one another.
The positions of places on the terrestrial sphere are described
by means of certain points and great circles on the sphere. There
are various pairs of circles which are used for reference; for
example, the equator (whose poles are the north and south poles
of the earth) and the meridian passing through the Koyal Observa-
tory at Greenwich, the equator and the meridian passing through
the observatory at Washington, etc. It will be observed that in
each case the reference circles are at right angles to each other, and,
accordingly, each of them passes through the poles of the other.
In an analogous manner the positions of bodies on the celestial
sphere are described by means of, or by reference to, certain points
and great circles on that sphere. There are four different systems
of circles of reference. As in the case of the terrestrial sphere,
each system consists of two circles, each of which passes through
the pole of the other, and, accordingly, is at right angles to the
other. Two of these systems are described in Arts. 70, 71, a third
in Art. 76, and the fourth in Art. 77. A point which will be referred
to in these systems is the north celestial pole. This is the point
where the earth's axis, if prolonged, would pierce the celestial
sphere. It is near the pole star, being about 1^° from it.
***... imagine the entire solar system as represented by a tiny circle the
size of the dot over this letter i." (Neptune the outermost planet known of
the solar system is 2700 millions of miles from the sun ; i.e. 30 times as far
as the earth.) " Even the sun itself, on this exceedingly reduced scale, could
not be detected with the most powerful microscope ever made. But on the
same scale the vast circle centred at the sun and reaching to Alpha Centauri
would be represented by the largest circle which could be drawn on the floor
of a room 10 feet square." (Todd, New Astronomy, p. 438.)
69-70.] CIRCLES ON THE CELESTIAL SPHERE. 87
70. The horizon system : Positions described by altitude and azi-
muth. For any place on the earth's surface, the point at which
the plumb line extended upwards meets the celestial sphere is
called the zenith ; the diametrically opposite point is called the
nadir. If a plane perpendicular to the plumb line be passed either
immediately beneath the observer's feet, or through the centre of
the earth, about 4000 miles below him, then the intersection of this
plane with the celestial sphere is called the horizon. (Since the
earth is so small and so far away from even the nearest star, two
parallel planes 4000 miles apart and passing through the earth
will appear, to a terrestrial observer, to intersect the celestial
sphere in the same great circle.)
Great circles passing through the zenith are perpendicular to the
horizon ; they are called vertical z (Zenith)
circles. The north point of the horizon
is the point which is directly north
from the observer. It is where the
vertical circle passing through the Soutt^(
north pole intersects the horizon.
This circle which passes through the
zenith and the pole is called the me-
ridian of the observer. The horizon
and the meridian are the reference circles in the horizon system.
The altitude (denoted by Ji) of a heavenly body is its angular
distance above the horizon. Thus the altitude of M (Fig. 53, in
which E is the earth and Z the zenith of the place of observation)
is Mm. The altitude of the zenith is 90°. The distance of a star
from the zenith is called its zenith distance ; this is obviously the
complement of the altitude.
The azimuth (denoted by A) of a heavenly body is the angle
between its vertical circle and the meridian. This angle is
measured usually along the horizon from the south point in the
direction of the west point, to the foot of the star's vertical circle.
Thus in Fig. 53 the azimuth of M is 180° + NZm, which is measured
by the arc 180° + Nm on the horizon.
NOTE. Any two points on the earth's surface have different zeniths.
Hence, the above system of describing positions on the celestial sphere is
peculiarly local. Moreover, a star rises in the eastern part of the horizon
88
SPHERICAL TRIGONOMETRY.
[CH. VII.
(altitude zero), mounts higher in the sky until it reaches the observer's merid-
ian, then sinks towards, and sets in, the west ; it is, accordingly, continually
changing its altitude and azimuth.
71. The equator system: Positions described by declination and
hour angle. The north celestial pole is the principal point of this
system. The celestial equator is the great circle of which that point
is the pole ; it is evidently the projection of the earth's equator
upon the^celestial sphere. The celestial equator and the meridian of
the observer are the reference circles in
the system now being described. In
Fig. 54, P is the north celestial pole, S
the south celestial pole, EQ, the celestial
equator ; also, HR is the horizon and Z
the zenith for some particular place on
the earth's surface. As said, in Art.
68, the stars move in parallel circles
whose axis is PS ; these circles are, accord-
ingly, parallel to the equator EQ,. The
angular distance of a star from the equator is called the declination
(denoted by D or 8) of the star ; north (or +) declination when the
star is north of the equator, and south (or — ) declination when the
star is south. Thus the declination of S^ is $3<V The angular dis-
tance of a star from the north pole is called its north polar distance ;
this is evidently the complement of the star's declination.'*
In 24 (sidereal) hours a star appears to make a complete revolu-
tion (i.e. to pass over 360°) about the celestial polar axis ; hence,
the star passes over 15° in 1 hour.-f The great circles passing
through the poles are called hour circles. Thus PSSS is the hour
circle of $3. The hour angle (denoted by //. JL.) of a star is the angle
between the meridian of the observer and the hour circle of the
Fia. 54
* The declination of the stars change by an exceedingly small amount in
the course of a year.
t The interval of time between two successive passages of the observer's
meridian by the sun (i.e. from noon to noon) is about 4 minutes longer than
the interval of time between two successive passages of the meridian by any
particular star. (This difference is due to the yearly revolution of the earth
about the sun. See text-books on astronomy.) The second interval is
called a sidereal day ; it is divided into 24 sidereal hours.
71-72.]
DECLINATION AND HOUR ANGLE.
star. This angle is measured towards the west. Thus, suppose
that a star is on the meridian at S4 ; its hour angle is then zero.
Twelve hours later the star will be at S0, and will have an hour
angle 180°. After a while it will be at Sl} just rising above the
horizon, and its hour angle will be 180° -|- S0PSl ; later it will be
at $3, having the hour angle 180° + SoPS3 ; later still it will be on
the meridian at S4, and its hour angle will be zero again. The hour
angle is usually reckoned in hours from 1 to 24, 1 hour being equal
to 15 degrees. Thus, when the star is at $0 its hour angle is 12 h.
The hour angle of a star is partly local; for only places on the same
meridian of longitude have the same celestial meridian. More-
over, the hour angle of a star is continually changing, and its
magnitude depends upon the time of observation. In Arts. 76,
77, the positions of stars are described in terms which are inde-
pendent of the time and place of observation.
In Arts. 73, 74, 75, the astronomical ideas so far obtained, are
used in the solution of two simple problems.
72. The altitude of the pole is equal to the latitude of the place of
observation. This theorem, which is necessary in Arts. 73, 74, is the
fundamental and most important theorem of spherical astronomy.
In Fig. 55, C represents the
centre of the earth, P its north
pole, and EQ its equator ; 0 is the
place of observation, say some place
in the northern hemisphere, Z is its
zenith and HR its horizon ; CPPl is
the celestial polar axis, Pl being the
north celestial pole. Draw OP2
parallel to CPlt P2 being on the
celestial sphere. The angle ROP2
is the altitude of the pole at 0,
since (see Arts. 68, 70) P1 and P2 are in the same direction from 0.
The latitude of a place is equal to the angle between the plumb
line and the plane of the equator. Thus, the latitude of 0 is equal
to OCE. Since OR and OP2 are respectively perpendicular to
CZ and CE, the angle jROP2= OCE; that is, the altitude of the
pole as observed at 0 is equal to the latitude of 0.
Fia. 55
00 SPHERICAL TRIGONOMETRY. [Cn.VII.
73. The time of day can be determined at any place ivhose latitude
is known, if the declination and the altitude of the sun at that time
and place are also known.
NOTE 1. The sun, unlike the stars, changes in declination from 23|° south
(about Dec. 22) to 231° north (about June 21), and then returns south. Its
declination is zero, that is, it is on the celestial equator, about March 20 and
Sept. 22. This change in declination is due to the revolution of the earth
about the sun, and to the fact that the plane of the earth's equator is inclined
about 23|° to the plane of its orbit about the sun. The latter plane is called
the plane of the ecliptic. The declination of the sun is given for each day
of the year in the Nautical Almanac. The altitude of the sun can be observed
with a sextant.
NOTE 2. The student should consult a text-book on astronomy for an
account of the special precautions and corrections necessary in connection
with this and similar astronomical problems.
In Fig. 56, P is the north celestial
pole, EQ the celestial equator, /S the
sun, and S0SSn is the small circle on
which the sun is moving at the given
time; Z is the zenith, and HR the
horizon, of the place of observation;
ZSM is . the sun's vertical circle, and
PSN is its hour circle.
^ ig midnight when the sun is at SQ,
and noon when the sun is at Sn. From
noon to noon is 24 hours. Hence, to find the time when the sun is
at S, determine the angle ZPS in hours (15° = 1 h.) ; subtract the
number of hours from 12, if it is forenoon; and add, if it is
afternoon.
Let I, h, D, respectively, denote the latitude of the place, and
the altitude and declination of the sun.
Then PR = I (Art. 72), SM= h, SN= D.
In ZPS, whose vertices are the sun, zenith, and pole,
Hence, the angle ZPS can be found.
73-74.] TO FIND THE TIME OF DAY. 91
EXAMPLES.
1. In New York (lat. 40° 43' N.) the sun's altitude is observed to be
30° 40'. What is the time of day, given that the sun's declination is 10° N.,
and the observation is made in the forenoon ?
2. In Montreal (lat. 45° 30' N.) at an afternoon observation the sun's
altitude is 26° 30'. Find the time of day, given that the sun's declination
is 8°S.
3. In London (lat. 51° 30'48"N.) at an afternoon observation the sun's
altitude is 15° 40'. Find the time of day, given that the sun's declination
is 12° S.
4. As in Ex. 2, given that the sun's declination is 18° N.
5. As in Ex. 3, given that the sun's declination is 22° N.
6. As in Ex. 1, given that the sun's declination is 10° S.
74. To find the time of sunrise at any place whose latitude is
knoivn, when the sun's declination is also known. This is a special
case of the preceding problem ; for at sunrise the sun is on the
horizon and its altitude is zero. The problem can also be solved
by means of the triangle RPSl (instead
of ZPS^ which is employed in Art 73).
For, in
= 90° - D, PR = I, PRSt = 90°.
= tan I tan D.
The angle RPSl (i.e. SoPS^ reduced
to hours, gives the time of sunrise (after
midnight). If ZPSl is found, then ZPS^ reduced to hours and
subtracted from 12 (noon), gives the time of sunrise. The time of
sunset is about as many hours after noon as. the tirno of sunrise
is before it.
In Fig. 57 the sun is north of the equator. When the sun is
south of the equator, PSl = 90° + D, and RPSl > 90° for places
in the northern hemisphere. The student can make the figure
and investigate this case, and also the case in which the place is
in the southern hemisphere.
92 SPHERICAL TEIGONOMETEY. [Cn. VII.
EXAMPLES.
Find the approximate time of sunrise at a place in latitude Z, when the
sun's declination is Z>, in the following cases :
1. I = 40° 43' N. (latitude of New York), D equal to : (a) 4° 30' N. (about
April 1); (6) 15° 10' N. (about May 1); (c) 23° N. (about June 10) ;'(<*) 5°N.
(about Sept. 10); (e) 6° S. (about Oct. 8); (/) 15° S. (about Nov. 3); (g) 23° S.
2. I = 51° 30' 48" N. (latitude of London), D as in Ex. 1.
3. I = 60° N. (latitude of St. Petersburg), D as in Ex. 1.
4. I = 70° 40' 7" N. (latitude of Hammerfest, Norway, D as in Ex. 1.
5. I = 29° 58' N. (latitude of New Orleans), D as in Ex. 1.
6. I = 33° 52' S. (latitude of Sydney, N. S. W.), D as in Ex. 1.
7. Find the approximate time of sunrise for other days and places.
75. Theorem. If the latitude of the place of observation is known,
then the declination and hour angle of a star can be determined from
its altitude and azimuth, and vice versa. For, in the triangle ZPS
(Fig. 56), ZP = 90° - I, SP= 90° - D, SZ= 90° - h, SPZ =
360° - H. A., PZS = A — 180°. Hence, if the latitude and any
two of the four quantities, viz., altitude, azimuth, declination,
hour angle, be known, then the remaining two can be found by
solving the triangle SPZ.
76. The equator system: Positions described by declination and
right ascension. In the system in Art. 71 the circles of reference
were the equator and the meridian of the observer. In the system
in this article the circles of reference are the equator and the circle
passinq throuqh the celestial poles and the
P (Pole) 1
vernal equinox. The vernal equinox is one
of the points where the ecliptic intersects
\ the equator ; namely, the point where the
sun, in its (apparent) yearly path among
the stars, crosses the equator in spring.
(See text-book on astronomy.) This point
may be called 'the Greenwich of the celestial
58 sphere. ( The ecliptic is the proj ection of the
plane of the earth's orbit on the celestial
sphere. The plane of the equator and the plane of the ecliptic are
inclined to each other at an angle of 23 -J-°. See Art. 73, Note 1.)
75-77.] DECLINATION AND RIGHT ASCENSION. 93
The right ascension (denoted by E.A.) of a heavenly body is
the angle at the north celestial pole between the hour circle of
the body and the hour circle of the vernal equinox. This angle
is measured from the latter circle towards the east, from 0° to 360°
or 1 h. to 24 h. ; it may be measured by the arc intercepted on the
equator. Declination has been defined in Art. 71.
In Fig. 58, P is the north celestial pole, E2Q the equator, E-f)
the ecliptic, and V the vernal equinox. If S is any star, then
for S D = $M) and R A = angle VPM= arc
77. The ecliptic system : Positions described' by latitude and longi-
tude. In this system the point and circles of reference are the
pole of the ecliptic, the ecliptic, and the great circle passing through
the pole of the ecliptic and the vernal equinox. The latitude of a
star is its angular (or arcual) distance
from the ecliptic; its longitude is the
angle at the pole of the ecliptic between
the circle passing through this pole and
the vernal equinox and the circle passing
through this pole and the star. This
angle may be measured by the arc inter-
cepted on the ecliptic. It is always
measured towards the east from the vernal FI(J r
equinox.
In Fig. 59, K is the pole of the ecliptic, E±C the ecliptic, P the
pole of the equator, E2Q the equator, and V the vernal equinox.
If S is any star, then
latitude of S = SM9 longitude of S = VKM= VM.
When the latitude and longitude of a star are known, its declination
and right ascension can be found, and vice versa. For, in the tri-
angle KPS (the triangle whose vertices are the star and the poles
of the equator and the ecliptic), KP=23±° (since QFC' = 23f>),
KS = 90°- lat., SKP = 90°- long., SP = 90°- D; SPK= VPK
-VPS = 90°- (360° - K.A.), if S is west of VP ; SPK = 90° +
R.A., if S is east of VP. If any two of these be known besides
KP, the remaining two can be found by solving KPS.
N.B. Questions and exercises on Chapter VII. will be found at page 109.
APPENDIX.
NOTE A.
ON THE FUNDAMENTAL FORMULAS OF SPHERICAL
TRIGONOMETRY.
1. The relations between the sides and angles of a right-angled spherical
triangle were obtained in Art. 26. The law of sines and the law of cosines
(Art. 36) for any spherical triangle have been derived by means of these
relations. (See Note 1, Art. 86.) These two laws can also be derived directly
by geometry -} this is done in Arts. 2, 3, below. Moreover, the law of sines
can be derived analytically from the law of cosines, as shown in Art. 4. In
Art. 5 it is shown how the relations for right-angled triangles can be derived
from these two laws. Other relations between the parts of a spherical tri-
angle have been referred to in Art. 40 ; these relations can also be deduced
by means of the law of cosines and the law of sines. The law of cosines is,
accordingly, the fundamental and most important formula in spherical trigo-
nometry.
2. Direct geometrical derivation of the law of cosines. Let 0-ABC
be a triedral angle, and ABC be the corresponding spherical triangle on
a sphere of radius OA. It is required to
find the cosine of the face angle COB, or,
what is the same thing, the cosine of the
side CB.
In OA take any point P, and through
P pass a plane MPN at right angles to
the line OA. Then OPN and 0PM are
right angles, and angle MPN = angle A.
Also, the measures (in degrees) of the
sides AB, BC, CA, are the same as the
measures of the face angles COB, BOA, AOC, respectively.
In MPN, MW2 = MP" + PN2 - 2 MP . PN cos MPN', (1)
in M ON, MN2 = MO2 + ON'2 - 2 M 0 • ON cos M ON. (2)
95
96
SPHERICAL TEIGONOMETET.
Hence, on equating these values of MN2 and transposing,
2 MO - ON cos MON = I/O2 - MP2 + ON2 - PN2 + 2 MP - PNcos MPN.
Now OM* - MP* = OP, and ON* - PN = OP, since OP3/ and
OP.V are right angles.
/. 2 MO - ON cos MON= 2 OP2 + 2 HP • PNcos MPN.
<.«.
cos a = cos & cos c + sin & sin c cos ^1.
(3)
Like formulas for cos 6, cos c, can be derived in a similar manner ; they
can also be written immediately, on paying regard to the symmetry in (3).
The formulas for cos A, cos JS, and cos <7, can be derived by means of the
polar triangle, as done in Art. 36, (7.
EXERCISES.
1. Make the figure and derive the law of cosines : («) when P is taken
at A ; (6) when P is taken in OA produced towards A.
2. Derive, the formula for cos 6 geometrically. (Take any point in OB,
and through this point pass a plane at right angles to OB. )
3. Derive the formula for cos c geometrically
3. Direct geometrical derivation of the law of sines. Let 0-ABC
be a triedral' angle, and ABC be the corresponding spherical triangle on
a sphere of radius OA. %
In OC take any point P, and draw PM
at right angles to the plane AOB, and in-
tersecting this plane in M. Through M
draw MG and MH, at right angles to OA
and OB respectively. Pass a plane through
the lines P2lf and MG.
Since PM is perpendicular to OAB, the
plane PMG is perpendicular to OA B (Euc.
XI. 18). Hence, since AGM is a right
angle, AGP is also a right angle. There-
fore angle PGM Bangle A. Similarly it
can be shown that angle PHM — angle B.
Fia...61
FM
PG OPsmAOC OP sin 6
. (1)
OP
APPENDIX. 97
= CO
.-. by (1) , (2) , sin A sin 5 = sin JS sin a.
sin_B
sin a sin b
In a similar way it can be shown that — — = sm . Hence
sin a sin c
sin A = sin# _ sinC,
sin a sin & sin c
Ex. 1. Show geometrically :
(a) that =
sin a sine sin 6 sine
Ex. 2. Make the derivation when M is not in the sector A OB.
4. Analytical derivation of the law of sines from the law of cosines.
cog A = cos a- cos b cos c. Art
sin & sin c
.-. 1 - cos2^ = 1 - / cos a -COB 6 cos cV»
\ sin b sin c )
_ sin2 b sin2 c — cos2 a — cos2 b cos2 c 4- 2 cos a cos & cos c .
sin2 6 sin2 c
_ (1 —cos2 6)(1 — cos2 c) — cos2 a— cos2 b cos2 c+2 cos a cos b cos c .
sin2 b sin2 c
. . 2 j _ 1 — cos2 a — cos2 b — cos2 c + 2 cos a cos 6 cos c
sin2 6 sin2 c
sin2J. _ 1 — cos2 a — cos2 & — cos2 c + 2 cos a cos & cos c ^
sin2 a sin2 a sin2 b sin2 c
Similarly, sm2 B and sm2 ^ can each be shown to be equal to the second
sin2 b sin2 c
member of (1). Hence,
sin A sin B sin C 2n ,
sin a sin b sin c sin a sin & sin c '
in which 2 w denotes the positive square root of the numerator of the second
member of (1),
Ex. 1. Show the truth of the statement made above.
Ex. 2. Show that the numerator in the second member of (1) is equal to
4 sin s sin (s — a) sin (s — 6) sin (s — c).
A A
SUGGESTION, sin A — 2 sin — cos — , and Art. 37, (4).
98 SPHERICAL TRIGONOMETRY.
5. Formulas for right-angled triangles derived from the general formulas.
In the triangle ABC let angle (7 = 90°. Then sin (7=1, and relations
(1), p. 45, become (2) and (2'), p. 30. Also, cos C = 0, and the third
formula in Art. 30, B becomes (1), p. 30. The three formulas in Art. 36, C
reduce to (5), (5'), and (0), p. 30, respectively. Formulas (3), (3'), (4)
and (4'), p. 30, can be derived from the others on that page. For
cos ^L = sin .B cos a [by (5')]=—-^^ [by (2'), (1)1 = — :
sin c cos 6 tan c
similarly,
tan c
Also,
[by (5')]=—— [by (2),
,
cosJ. sin B cos a sin 6 cos a sin b
similarly, tan_B = ^L&.
sin a
Other relations in triangles (see Art. 40) can also be used in the derivation
of the formulas for right-angled triangles.
EXERCISES.
1. Deduce the law of cosines : (1) directly, by geometry ; (2) by means
of the relations in a right-angled triangle.
2. Deduce the law of sines : (1) analytically, from the law of cosines
(2) directly, by geometry ; (3) by means of the relations in a right-angled
triangle.
3. Deduce the ten relations between the sides and angles of a right-angled
spherical triangle : (1) by means of the relations between the sides and angles
of the general spherical triangle ; (2) directly, by geometry.
NOTE B.
[Supplementary to Art. 58.]
DERIVATION OF FORMULAS FOR THE SPHERICAL EXCESS
OF A TRIANGLE.
J. Cagnoli's Formula. (In terms of the sides.}
BW$E = *m%(A + B + (7-180°) = - cos$(A+ B + (7)
'== sin l(A + B) sin \C - cos \(A + B) cos \C
= sintCcos?C[cos A(« - &)- cos i(« + &)] [Art. 39, ^1), (3)]
cos \ c sin a sin b
n ^
2 cos £ a cos £ b cos £ c
2n [Note A, Art. 4, Eq. (2)]
APPENDIX.' ' 99
II. Lhuillier's Formula. (In terms of the sides.)
sin KA + B + 0-180°)
0
gin ^ + J)- Bin 1080° -C) . [Plane m 94]
cos K^ + -B) + cosK180° - C)
sin %(A + B} - cos
sin KS — 6) sin ^(s — a) / sin s sin (a — c)
— '
cos £ s cos \ (s — c) * sin (s — a) sin (s — &)
[Art. 37, (6); P/a»e TVty., p. 94]
= Vtan £ s tan ^(s — a) tan ^(s — &) tan ^(s — c).
III. Formula in terms of two sides and their included angle.
cos \E = cos %(A + B + C - 180°) = sin %(A + B + <7)
= cos ^(-4 + 5) sin $C + sin £(.4 -f- 5) cos ^ (7
= [cos Ka + 6)sin2^O+ cos^(a - 6) cos2 JC.] sec ^c
[Art. 39, (1), (3)]
= (cos £ ci.cos ^ & + sin $ a sin $ 6 cos (7) sec £ c. (2)
Hence, from (1) and (2), on division and reduction,
tanAJ?= tan |o tan ^6 sinC ,
1 + tan \ a tan \ b cos C
On taking the reciprocals and reducing, this takes the form
. . „ cot i a cot \ b + cos (7
CO I * Hi = * — •
sinC
•
ESTIONS AND EXERCISES FOR PRACTICE
AND REVIEW.
CHAPTER I.
1. On a sphere let N be the pole of a great circle ABC, and P be any
point on the surface between N and ABC; also let DPNG be a semicircle
drawn through P at right angles to AB C, and let it intersect ABC in D
and Cr : prove (a) that PD is the shortest great-circle arc that can be drawn
from P to ABC ; (5) that PNG is the longest great-circle arc that can be
drawn from P to ABC.
2. Show that the greater the distance of the plane of a small circle from
the centre of the sphere, the less is the circle.
3. The radius of a sphere is 10 inches, and the radius of a small circle
upon it is 6 inches. Find : (a) the distance between the centre of the sphere
and the centre of the small circle ; (&) the angular radius of the small circle ;
(c) the polar distance (or arcual radius) of the small circle ; (cf) the dis-
tance , on the sphere from the small circle to the great circle having the
same axis.
4. Prove that if a spherical triangle has two right angles, the sides oppo-
site them are quadrants, and the third angle has the same measure as its
opposite side.
5. Prove that in any spherical right triangle an angle and its opposite
side are always in the same quadrant.
6. Prove that any side of a spherical triangle is greater than the difference
between the other two sides.
7. Prove that each angle of a spherical triangle is greater than the differ-
ence between 180° and the sum of the other two angles.
8. Show that the surface of a sphere is eight times the surface of a trirec-
tangular triangle:
9. (a) Show that a trirectangular triangle is its own polar; (&) show
that a triquadrantal triangle is its own polar.
10. Show that if two great circles are equally inclined to a third, their
poles are equidistant from the pole of the third.
101
102 SPHERICAL TRIGONOMETRY.
11. Show that the arc through the poles of two great c .cles cuts both
circles at right angles.
12. A ship sails along the parallel of 45° N. a distance of 600 nautical
miles. Find the difference of longitude that she has made.
13. Two places in latitude 60° N. are 150 statute miles apart. Find their
difference of longitude. [Take the radius of the earth as 3960 miles.]
14. Compare the lengths of the parallels of 30° N., 45° N., and 60° N.,
with the length of the equator.
15. Prove that if the first of two spherical triangles is the polar triangle
of the second, then the second is the polar triangle of the first.
16. Show that in two polar triangles each angle of the one is the supple-
ment of the side opposite to it in the other.
17. Show that the sum of the angles of a spherical triangle is greater than
two, and less than six, right angles.
18. Discuss the following cases, in which A, a, and b are given in a spheri-
cal triangle ABC:
I. ^1 = 90°: (1) & = 90°; (2) &<90°(a<6, a = b, a>b and <ir-b,
a = TT - b, a>7r-&); (3) b >90°(a<7r - 6, a = TT - b, a>tr-b and
<6, a = b, «>&).
II. ^1<900: (1) b = W°(a<A, a = A, a> A and < 6, a = b = 90°,
a > 6) ; (2) b < 90°(a <p, a =j>, a >p and <b, a = b, a>b and <TT - 6,
a = TT — b, a>?r — 6); (3) £>>90°(a<p, a = p, a>p and <TT — 6,
a — TT — b, a > TT — b and < b, a = b, a > 6) . [For definition of p, see p. 26.]
III. ^4>90°: (1) & = 90°(a = &, a between & and TT - ft, a between
TT — b and ^?) ; (2) &<90°(a>p, a=.p, a<j» and > &, a = 6, a between 6
and TT — 6) ; (3) 6>90°(«<&, a>6 and <j9, a<p and >?r — 6,
a between 6 and TT — 6, a = 6).
CHAPTER II.
1. Define spherical angle, spherical triangle, Napier's circular parts,
polar triangle, quadrantal triangle, oblique spherical triangle, pole of an
arc, spherical excess, spherical polygon.
2. In a right-angled spherical triangle show that : (a) It is impossible
for only one of the three sides to be greater than 90° ; (6) The hypote-
nuse is less than 90° only when both the other sides are in the same
quadrant ; (c) If another part besides the right angle be right, the triangle
is biquadrantal.
3. Prove, by geometry and by trigonometry, that in a right spherical
triangle an angle and its opposite side are always in the same quadrant, that
is, either both are less or both are greater than 90°.
QUESTIONS AND EXERCISES. 103
4. Trove tbiat in a right spherical triangle ABC, (C - 90°) : (a) sin A =
cos B + cos 6; (6) cos c = cot A cot B ; (c) cos c = cos a cos b.
5. (a) Mention in order Napier's circular parts, and state the two prin-
cipal rulesr for their use. (&) State Napier's Kules and write the ten for-
mulas for the right spherical triangle by means of them, (c) Prove three
of these formulas.
6. What formulas should be used to find B, a, and & of a right spherical
triangle ABC (C = 90°) when A and c are given ? What formula includes
all the required parts ?
7. Show how to obtain the formulas for finding a, B, and C of a qua-
drantal triangle, when A and 6 are given and c = 90°.
8. Given one side and the hypotenuse of a right spherical triangle, write
all the formulas for the solution and check, and state how the species of each
part will be determined.
9. How many solutions are there for a right spherical triangle ABC,
given side b and angle B ? Discuss fully.
10. Given A and b of a right spherical triangle ABC (<7= 90°) : write
and derive formulas for computing each of the parts B, a, and c in terms of
A and 6 only ; also the check formula.
11. Show how to solve a right spherical triangle, having given (a) the
sides about the right angle ; (&) the two oblique angles.
12. (a) Show how the solution of a quadrantal triangle may be reduced
to that of a right triangle. (&) Write the relations between the sides and
angles of a quadrantal triangle ABC, in which c = 90°.
13. In a spherical triangle ABC, A = B : write the relations between the
sides and angles of ABC.
14. If A be one of the base angles of an isosceles spherical triangle whose
vertical angle is 90° and a the opposite side, prove that cos a = cot A ; and
determine the limits within which it is necessary that A must lie.
15. Show how oblique spherical triangles can be solved by means of right
spherical triangles. (Six cases.)
16. In a right spherical triangle ABC (C = 90°) prove that: («) sin2 B
— cos2 A — sin2 b sin2 A; (&) sin A sin 2 b — sin c sin 2 B ; (c) sin 2 A sin c =
sin 2 a sin B ; (d) sin 2 a sin 2 b = 4 cos A cos B sin2 c ; (e) cos2 A sin2 c =
sin'2 c - sin2 a ; (/) sin2 A cos2 c = sin2 .4 - sin2 a.
17. (a) In ABC, if C = 90°, and a = b = c, prove that sec A = I + sec «,
(6) In ABC (C= 90°) show that if b = c = -, then cos a = cos A.
2
18. In a right spherical triangle whose oblique angles are 72° 34' and
59° 42', find the length of the perpendicular from the right angle upon the
base, and the angles which it forms with the sides.
104 SPHERICAL TRIGONOMETRY.
19. Two planes intersecting at right angles are intersected by a third
plane making with them angles of 60° and 75° respectively. Find the angles
which the three lines of intersection make with each other.
20. Two planes intersect at right angles ; from any point of their line of
intersection one line is drawn in each plane making the respective angles 60°
and 73° with the line of intersection. Find the angle between the two lines
thus drawn.
21. A triangle whose sides are 40°, 90°, and 125° respectively, is drawn on
the surface of a sphere whose radius is 8 feet. Find in feet the length of
each side of this triangle, and also the angles of the polar triangle. Write
the formula for finding either angle in terms of functions of the sides.
22. Solve the following spherical triangles given: (1) Right triangle,
hypotenuse = 140°, one side = 20°. (2) Sides 90°, 50°, 50°. (3) Sides 100°,
50°, 60°. (4) Sides each 30° in length. (5) A = 100°, C = 90°, a = 112°.
(6) .4 = 80°, a = 90°, ft = 37°. (7) a = b = 119°, (7=85°. (8) Triangle
PQR, R = 90°, P= 63° 42', Q = 123° 18'. (9) Right triangle, one angle =
110° 30' 20", hypotenuse = 75° 45'. (10) A = 90°, ft = 21° 30', c = 122° 18'.
.(11) 5 = 90°, C= 79° 40', b =137° 52'. (12) .4=90°, « = 108°23, c=37°42'.
(13) #=90°, ,4=43° 10', a = 78°35'. (14) .5=90°, C =33° 57', .4=43° 18'.
(15) ,4 = 87° 40' 20", b = 33° 42' 40", 5 = 90°. (16) .4 = 33° 42' 40", 6 =
87° 40' 20", .8 = 90°.
CHAPTER III.
1. In a spherical triangle ABC prove that : (a) sin a : sin yli=sin b : sin B
= sin c : sin C ; (ft) cos a = cos ft cos c + sin ft sin c cos A ; (c) cos A =
— cos B cos C+ sin B sin C cos a ; (d) cos %A= Vsin s sin(s— a) -f-sin ft sin c,
where s = £(a + ft + c) ; (e) tan \ A cot \ B = sin (s — ft) cosec (s — a).
2. Give the equations (or proportions) known as Napier's Analogies.
Derive them.
3. Derive formulas giving the values of sin A, cos A, tan A, and cos c, in
terms of functions of a, ft, and c.
4. In a spherical triangle ABC show that : (a) If a = b = c, then sec^l
= 1 + sec a. (6) If ft -f c = 180°, then sin 2 B + sin 2 C = 0. (c) If C= 90°,
then tan i(c + a) tan J(c - a) = tan2 1.
2
5. In an equilateral spherical triangle show that : (a) 2 sin — cos | = 1,
Zt -j
and hence, that such a triangle can never have its angle less than 60°, nor its
side greater than 120° ; (ft) 2 cos .4 = 1 - tan2 -•
6. Show that : (a) If the three angles of spherical triangle ABC are
together equal to four right angles, then cos2 ^ = cot A cot B. (6) If x is
2
QUESTIONS AND EXERCISES. 105
the side of a spherical triangle formed by joining the middle points of the
equilateral triangle of side a, then 2 sin - = tan -•
2 2
7. (a) In a spherical triangle ABC show that, if b -f c = 90°, then
cos a = sin 2 c cos2 — . (6) If a be the side of an equilateral triangle and a'
2
that of its polar triangle, prove cos a cos a' = ±.
8. (a) If, in a triangle ABC, I be the length of the arc joining the middle
point of the side c to the opposite vertex C, shov^that cos I = (cos a + cos 6)
-r- 2 cos-. (6) In a right spherical triangle ABC (C = 90°), if a, ft be the
arcs drawn from C respectively perpendicular to and bisecting the hypote-
nuse c, show that sin2 - (1 + sin2 a) = sin2 ft.
9. (a) Prove that the half sum of two sides of any spherical triangle is in
the same quadrant as the half sum of the opposite angles. ^6) Two sides of
a spherical triangle are given : prove that the angle opposite the smaller of
them will be greatest when that opposite the larger is a right angle.
10. ABC is a spherical triangle of which each side is a quadrant, and Pis
a point within it. Prove that cos2 PA + cos2 PB + cos2 PC = 1.
11. In a spherical triangle, if A = 36°, B = 60°, and C = 90°, show that
a + 6 + c = 90°.
CHAPTER IV.
1. (a) Name the six cases for solution of spherical triangles. (6) Dis-
cuss each case in detail, writing the formulas used in the solution, and deriv-
ing these formulas, (c) Solve an example under each case. Test the result
by (1) solving by right triangles, (2) solving without logarithms, (3) using a
check formula.
2. How many solutions are possible for the oblique spherical triangle
ABC, given A, B, and a ? Discuss in full the question of one solution, two
solutions, or no solution. Plan the solution.
3. In a spherical triangle ABC, two sides a and b and the included angle
C are given. Write all the formulas used in the solution and check ; describe
fully the process of solution. Derive the formulas used.
4. Write and deduce the formulas for rinding A, B, and C of any spheri-
cal triangle when a, b, and c are given.
5. Given A, _B, and C. Show how to find the remaining parts, writing
the formulas to be used.
6. In an equilateral spherical triangle the side a is given. Find the
angle A.
106 SPHERICAL TRIGONOMETRY.
7. Solve the spherical triangle whose sides are 70°, 60°, and 50°. Solve
the plane triangle obtained by connecting by straight lines the vertices of
this spherical triangle, the sphere on which it is drawn being 2 feet in
diameter.
8. In a triangle ABC on the earth's surface (supposed spherical) a = 483
miles, b = 321 miles, C = 38° 21'. Find the length of the side c. [Earth's
radius = 3960 miles.]
9. Two planes intersect at an angle of 75°. From any point of their line
of intersection one line is drawn in each plane, making the respective angles
55° and 80° with the line of intersection. Find the angle between the lines
thus drawn.
10. Two planes intersecting at an angle of 65° are intersected by a third
plane, making with them the respective angles 55° and 82°. Find the angles
which the three lines of intersection make with one another.
11. A solid angle is contained by three plane angles 62°, 83°, 38°. Find
the angle between the planes of the angles 62° and 38°.
12. Two of the three angles which contain a solid angle are 42° and 65° 30',
and their planes are inclined at an angle of 50°. Find the angle of the third
plane face and the angles at which this third plane is inclined to the other
two planes.
13. A pyramid has each of its slant sides and base an equilateral triangle.
Find the angle between any two faces.
14. A pyramid each of whose slant faces is an equilateral triangle has a
square base. Find the angle between any two slant faces, also the angle
between any slant face and the base.
15. In the following cases ABC is a three-sided spherical figure each of
whose sides is an arc of a great circle. Select those which are spherical tri-
angles, and give reasons for so doing. Explain why the other figures can-
not be triangles. Solve the triangles and check the results. (Solve some
without using logarithms.)
(1) a = 76°, b = 54°, c = 36°. (2) A = 54° 35' 20", b = 104° 25' 45",
c = 92°10'. (3) A = 107-° 47' 7", B = 38° 58' 27", c= 51° 41' 14".
(4) A = 60°, B = 80°, C = 100°. (5) A = 120°, B = 130°, C = 80°.
(6) .4=54° 35', 6 = 104° 24', c=95° 10'. (7) .4=61° 37' 53", B= 139° 54' 34",
6 = 150° 17' 26". (8) a = 72°18', b =146° 35', c=98°ll'. (9) .4 = 125° 15',
C= 85° 12', 6 = 100°. (10) .4=50°, 5=114° 5' 8", 6 = 50°. (11) ^1=83° 40',
b = 73° 45', a = 30° 24'. (12) A = 83° 40', b = 30° 24', a = 73° 45'.
(13) A = 97° 20', a = 94° 37', b = 36° 17'. (14) a = 127° 40', b = 143° 50',
c = 139° 39*. (15) A = 40°, B = 30°, C = 20°. (16) A = 40° 35',
B = 36° 42', c = 47°18'.
QUESTIONS AND EXERCISES. 107
CHAPTER V.
[In the following exercises,
w = Vsin s sin (s — a) sin (s — &) sin (s — c),
and N = V- cos #cos (S - -4) cos (£- B} cos (# - C) ;
also, 7% ra, 7*5 rc, denote the radii of the circles inscribed in the spherical tri-
angle ABC and its three colunar triangles, and jf?, _Z?a, Rb, Rc denote the
radii of the circumscribing circles of these triangles.]
1. Given a spherical triangle. AB C, find (1) the radius of the inscribed
circle ; (2) the radius of the circumscribing circle ; (3) the radii of the
inscribed circles of the colunar triangles ; (4) the radii of the circumscribing
circles of the colunar triangles.
Show that :
sins
N
3.
cos (S - A) cos (S - B) cos (# - C)
4. (a) Cot R cot Ra cot Rb cot Rc = N2 ;
(6) Tan R cot Ra cot Rb cot Jf?c = cos2 S.
— cos $ sin s
5. Tan .R = 4 tan r
sin a sin & sin c sin A sin B sin (7
6. Tan ra tan r& tan rc = tan r sin2 s.
7. Tan R -f cot r = tan JKa -f cot ra = tan J?5 + cot rb
= tan Rc + cot rc = | (cot r -f cot ra + cot r& + cot rc).
8. Tan^tanr = -COS^sina^-cos-ysln6^etc. Write the other for-
mulaofthisset. sins sin 5
9. Tan2 j£ + tan2 Ra + tan2 Rb + tan2 J?c = cot2 r + cot2 ra + cot2 rb + cot2 rc.
10. Tan r tan ra tan r6 tan rc = w2 ; cot r tan ra tan rt tan rc = sin2 s.
11. In any equilateral triangle, tan R = 2 tan r.
12. TanJgB= tan^ = cos(^-^)^ sin^a
cos jS N sin ^. sin \ b sin ^ c
_ &) + gin(s ^
7i 2 71
Write the corresponding formulas for Rb and 7?c.
13. Cot ra + cot r& + cot re — cot r = 2 tan .R.
14. Find the radii of the circles connected with some of the triangles -in
Ex. 15 of the preceding set.
108 SPHERICAL TRIGONOMETRY.
CHAPTER VI.
1. Define the following terms : zone of a sphere, lune, spherical degree,
spherical excess of a triangle, spherical excess of a (non-re-entrant) polygon,
spherical excess of any figure on a sphere, spherical measure and spherical
degree measure of a solid angle, spherical pyramid, spherical sector, spherical
segment.
2. Derive the area of the surface of a sphere.
3. Derive the area of a spherical triangle.
4. Discuss fully the measurement of solid angles.
5. Show how to find the spherical excess of a figure on a sphere when
the area of the figure is given (in square units).
6. State and deduce Roy^s Rule for computing the spherical excess of a
triangle of known area on the earth's surface.
7. Derive the volumes of a sphere, a spherical pyramid, a spherical
sector, and a spherical segment.
8. The area of an equilateral triangle is one-fourth the area of the
sphere : find its sides and angles.
9. If the three sides of a spherical triangle measured on the earth's
surface be 12, 16, and 18 miles, find the spherical excess.
10. If a = b and C = - show that tan#° = sm2 a . (jn ABC.)
2' 2 cos a
11. If a = b = 60° and c = 90°, show that E = cos-1 $. (In ABO.)
12. If C = 90° in ABC, then E = 2 tan-1 (tan £ a tan \ 6).
13. In a triangle on the earth's surface (assumed spherical), two sides
are 483 and 321 miles, and the angle between them is 38° 21'. Find the area
of the triangle in square miles. [Radius of earth = 3960 miles.]
14. The sides of a triangle on the earth's surface (supposed spherical)
are 321, 287, and 412 miles ; find the area.
15. Prove that in a right triangle ABC (C = 90°),
cosiff = cos*qcosa6, and 8iniJr = ggJg8in'*6.
cos \c cos I c
16. The spherical excess of a triangle on the earth's surface is 2". 6.
Find its area, the radius of the earth being taken as 3960 miles.
17. Find the fraction of the earth's surface (supposed spherical) con-
tained by great-circle arcs joining London, New York, and Paris. Find the
spherical degree measure, and the spherical measure of the angle subtended
at the centre of the earth by this part of the earth's surface.
QUESTIONS AND EXERCISES. 109
18. Find the spherical excess of some of the triangles in Ex. 15, p. 104.
Also find their areas in square inches on spheres of radii, say, 4 inches,
10 inches, 12 inches, 20 inches, a inches.
19. Find the spherical measures and the spherical degree measures of the
solid angles corresponding to the triangles taken in Ex. 18.
CHAPTER VII.
1. Given the latitude and longitude of each of two places : show how to
find the shortest distance between these places, and the direction of one place
from the other.
2. Given the latitudes and longitudes of three places on the earth's sur-
face, and also the radius of the earth : show how to find the area of the
spherical triangle formed by arcs of great circles passing through them.
3. Given the sun's altitude and declination and the latitude of a place :
show clearly how the time of day may be determined.
4. If d represents the sun's declination, what formulas will be required in
order to determine the time of sunrise for a place whose latitude is I ?
5. Show what formulas must be used to find the length of a degree of
longitude on the earth's surface for a place whose latitude is Z, r representing
the radius of the earth.
6. The shortest distance d between two places and their latitudes I and I'
are known ; find their difference of longitude.
7. Given the obliquity of the ecliptic w, and the sun's longitude X, show
that if a and d denote his right ascension and declination respectively, then
tan a = cos a> tan X, and sin S = sin w sin X.
8. The faces of a regular dodecaedron are regular pentagons, three faces
meeting at each vertex. Find the diedral angle at the edge of the solid.
9. The ridges of two gable roofs meet at right angles ; each roof is
inclined to the horizontal at an angle of 65°. Find the diedral angle between
the planes of the two roofs, and the angle their line of intersection makes
with the ridge of either roof.
10. What is the direction of a wall in latitude 52° 30' N. which casts no
shadow at 6 A.M. on the longest day of the year ?
11. Two ports are in the same parallel of latitude, their common latitude
being J, arid their difference of longitude 2 X. Show that the saving of dis-
tance in sailing from one to the other on the great circle instead of sailing
due east or west, is
2 r {X cos I — sin~1(sin X cos Z)},
\ being expressed in radian measure, and r being the radius of the earth.
110 SPHERICAL TRIGONOMETRY.
12. If a ship sails from New York (40° 28' N. , 74° 8' W.) starting due east,
and continues her course on an arc of a great circle, what will be her lati-
tude when she reaches the meridian of Greenwich, and in what direction will
she then be sailing ?
13. Find the distance between New York (40° 28' N., 74° 8' W.) and Cape
Clear (51° 26' N., 9° 29' W.), and the bearing of each from the other. [Radius
of earth = 3960 miles.]
14. From Victoria, B.C. (48°25'N., 123° 23' W.), a ship sails on an arc
of a great circle for 1250 miles, starting in the direction S. 47° 35' W. Find
its latitude and longitude, taking the length of 1° as 69| miles.
15. Two places are both in latitude 50° N., and the difference of their
longitudes is 60°. Find the distance between them (a) along the parallel of
latitude, (&) along a straight line, (c) along a great circle. [Earth's radius
= 3960 miles.]
16. What will be the first course and the shortest (great circle) distance
passed over in sailing from a place in latitude 43° N. to another place 80°
east of it and in the same latitude ? What is the distance between the two
places along the parallel ? What is the straight-line distance between them ?
17. At what hours will the sun rise in London (51° 30' 48" N.) and New
York (40° 43' N.) when its decimation is respectively 23° N., 20° N., 15° N.,
10° N., 5° N., 5° S., 10° S., 15° S., 20° S., 23° S. ?
18. When the sun's declination is 18°, find his right ascension and
longitude.
19. What is the altitude of the sun above the horizon when its angular
distance from the south point is 75° and from the west point is 60° ?
20. The right ascension of Sirius is 6h 38m 37S.6, and his declination is
16° 31' 2" S. ; the right ascension of Aldebaran is 4h 27m 258.9, and his decli-
nation is 16° 12' 27" N. Find the angular distance between these stars.
21. If the sun's declination be 20° 45' N. and his altitude be 41° 10' at
3 P.M., find the observer's latitude.
22. What will be the altitude of the sun at 3.30 P.M. in San Francisco
(37° 48' N.), its declination being 15° S. ?
23. In Bombay (18° 54' N.) the altitude of the sun is observed to be
27° 40'. If the sun's declination is 7° S. and the observation is made in the
morning, find the hour of the day.
24. Find the latitude and longitude of a star whose right ascension is
4h 40™, and declination 57°.
25. Find the distance in degrees between the sun and moon when their
right ascensions are respectively 15h 12', 4h 45', and their declinations are
21° 30' S., 5°30'N.
QUESTIONS AND EXERCISES. Ill
26. Find the length of the longest day in the year at the following places
(the sun's greatest declination being 23° 27' N.) : London (51° 30' 48" N.),
New York (40° 43' N.), Montreal (45° 30' N.), St. Petersburg (60° N.), Hong
Kong (22°17'N.).
27. Find the length of the shortest day in the year at the places mentioned
in Ex. 26. (The sun's declination is then 23° 27' S.)
28. At Copenhagen (55° 40' N.), at an afternoon observation, the sun's
altitude is 44° 20' ; find the time of day, the sun's declination being 18° 25' N.
29. At what time of day will the sun have an altitude of 53° 40' for a
place in latitude 40° 35' N., his declination being 13° 48' N. ?
30. What will be the sun's altitude at 3.30 P.M. at a place in latitude
44° 40' N., his declination being 18° N. ?
31. What will be the sun's altitude at 10 A.M. at a place in latitude
44° 40' N., his decimation being 18° S. ?
32. What is the sun's declination when his altitude at a place in latitude
37°48'N. is 25° at 4 P.M. ?
NOTE. The Spherical Trigonometries of M'Clelland and Preston, Casey,
and Bowser, contain especially good collections of exercises. See Art. 40,
ANSWERS TO THE EXAMPLES.
CHAPTER I.
Art. 24. I. 4. A = 88° 12.2', B = 74° 34.7', C = 43° 8' ; A = 118° 33.2',
# = 113° 11. 2', C=92°45'. II. 4. a = 72° 40.6', 6=67° 45.8', c = 51° 43.1';
a=71°22.7', 6 = 108° 37.3', c = 104°56.7'. III. 4. ^L=63°56', £=126° 21.2',
c = 77°3'; £ = 32° 47.1', 0= 62° 30.7', a = 84° 29.5'. IV. 5. 6 = 70° 5.7',
c = 102°51.3', ^1 = 68° 35.8'; a = 46° 1.5', c = 86° 0.7', B = 122° 55.8'.
VI. 3. .B = 59° 40.1', <7=114°55', c = 96° 31.1', and B = 120° 19.9',
C = 27° 49.6', c = 30° 45.4'; J? = 65° 1.8', C = 97° 16.9', c = 100° 26';
C= 110° 43.1', b = 33° 8.6', c = 60° 28.8'; C = 165° 3.3', 6 = 125° 1.7',
c = 162° 55.7', and C = 119° 47', c = 81° 7', b = 54° 58.3'.
CHAPTER II.
Art. 27. 4. c = 82° 33.9', ^ = 60° 51.2', B = 76° 56.1'. 5. a=33°0.25',
b = 36° 29.4', c=47£37. 8'.
Art. 31. 5. (!>#*= 86° 30.9', .4=36° 30.2', 5=87° 25.4'. (2) 6 = 138° 24.4',
4 = 68° 41.9', B =129° 43.1'. &*) a = 35° 50.6', 6 ^.75° 39. 5', B = Sl° 29.1'.
(tf a = 42° 49.8', b = 27° 47.3', c = 49° 33'. &£b = 33° 37.4', c = 79° 2',
_B=34°20.1'; and 6 = 146° 22.6', c=100°58', J?=145° 39.9'. (J6) a =35° 16.4',
c = 51° 10.8', 5 = 55° 18.6'.>
Art. 32. 1. (1) fc = 54°20', ^=32°0.75', J? = 57°59.25', (7=93°59.3';
(2) 6 = 66° 29', c = lll°29.4', jB=50°17', O=128°41.2'. 2. (1) 6 = 59° 56.2',
A = 130°, B = 52° 55.5'. (2) a = 135° 33', b = 100° 58.6', C = 101° 24.7'.
CHAPTER III.
Art. 37. I. 2. ^1=55° 58.4', J5=74° 14.6', (7=103° 36.6'. 3. A =43° 58',
B = 58° 14.4', G = 108° 4.8'. II. 3. a = 39° 29.6', b = 35° 36.2', c = 27° 59'.
4. a = 130° 49.6', 6 = 120° 17.5', c = 54° 56.1'.
CHAPTER IV.
Art. 42. 2. A = 41° 27', B = 66° 26.4', G = 106° 3,2'. 3. A = 144° 26.6',
^ = 26° 9.1', C= 36° 34. 7'.'
Art. 43. 1. a = 43° 36', b = 41° 20.9',' c = 33° 7.4'. 2. a = 111° 40.2',
* = 91° 17.2', c = 71° 7.4'.
113
114 ANSWERS TO THE EXAMPLES.
Art. 44. 2. A = 101° 21.2', B = 54° 57.9', c = 79° 9.5'. 3. B = 78° 20.6',
<C = 47°47', a = 82° 42'.
Art. 45. 1. a = 63° 15.1', b = 43° 53.7', C= 95° 1'. 2. 6 = 86° 39.5',
c = 68° 39.5', A = 59° 44'.
Art. 46. 2. B = 36° 35.5', C = 51° 59.7', c = 42° 38.9'. 3. B = 59° 3.5',
C = 97° 38.8', c = 56° 56.9' ; B = 120° 56.5', C = 28° 5.2', c = 23° 27.8'.
Art. 47. 1.6 = 154° 45.1', c = 34° 9.1', C = 70° 17.5'. 2. ^1 = 164° 43.7',
a = 162° 37.5', c = 124° 40.6' ; A = 119° 18.7', a = 81° 18.7', c = 55° 19.4'.
CHAPTEE VI.
Art. 53. 1. 2827.44 sq. in. 2. 392.7 sq. in. 3. 8.25 sq. ft.
Art. 55. 1. 1.396 sq.ft. 2. 64.14 sq. ft.
Art. 56. 24° 37' 47" (.42986), 33° 56.6' (.59213), 27° 10.4' (.47426),
12° (.20944), 86° 20' (1.5068), etc.
Art. 57. 1. 42.986 sq. ft., 59.213 sq. ft., 47.426 sq. ft. 2. 130.9 sq. in.,
941.75 sq. in.
Art. 61. 1. Spherical degree measure = 12, spherical measure = .20944,
2. Spherical degree measure = 24.63, spherical measure = .42986.
Art. 64. 1. 143.29 cu. ft., 197.38 cu. ft., 158.09 cu. ft., 1090.8 cu. in.,
7847.9 cu. in., etc. 2. (a) 1357.17 cu. in. (6) 904.78 cu. ft.
CHAPTER VII.
Art. 66. 2. 8° 4.3' S. ; course, S. 45° 6 E. 5. (a) On the equator in
long. 18° 56' E. ; course, S. 47° 39' E. (6) Lat. 42° 21' S., long. 108° 56' E. ;
course, E. (c) On the equator in long. 161° 4' W. ; course, N. 47° 39' E.
6. Distance =(51° 19.8')= 3547.675 mi. ; bearing of New York from Liver-
pool is N. 71° 6.8' W., and bearing of Liverpool from New York is N. 48° 5.8' E. ;
lat. 51° 441' N. ; course, N. 65° 38' E.
Art. 73. 1. 8.08A.M. 2. 2.33P.M. 3. 2.59 P.M. 4. 4.09P.M.
5. 6.09 P.M. 6. 9.46 A.M.
Art. 74. 1. (a) 5.44; (6) 5.06; (c) 4.34; (<*) 5.43; (e) 6.21;
(/) 6.53; (0) 7.26. 2. (a) 5.37; (6) 4.40; (c) 3.51; (d) 5.35;
(e) 6.30; (/) 7.19; (0) 8.09. . 3. (a) 5.29; (6) 4.08; (c) 2.51;
(d) 5.25; (e) 6.42; (/) 7.51. (j)
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