Download the elements of spherical trigonometry

THE
ELEMENTS
OF
SPHERICAL
TRIGONOMETRY.
BY JAMES HANN,
J,\TE MATHEMATICAL MASTER OF KING'S COLLEGE 8CHOOL, LONDON.
gi |Ufo ^bition, mrb Corwcttb,
BY CHAELES H. BOWLING, C.E.,
LATE OF TBINITY COLLEGE, DUBLIN.
LONDON:
VIRTUE BROTHERS & CO., 26, IVY LANE,
v
'
'
PATEItNOSTEIt HOW.
18GG.
PREFACE
TO THE
SECOND EDITION.
The first edition of Mr. Hann's " Elements of Spherical
Trigonometry" has been carefully revised, and corrected,
and several practical problems from the prize questions of
different universities have been added to the work. The
author states, in his preface to the former edition:—
"In the compilation of this work, the most esteemed
writers, both English and foreign, have been consulted, but
those most used are De "Fourcy and Legendre.
"Napier's ' Circular Parts' have been treated in a manner
somewhat different to most modern .writers. The terms
conjunct and adjunct, used by- Kelly and others, are here
retained, as they appear to be more conformable to the
practical views of Napier himself."
This edition is recommended with increased confidence
to the judgment of all mathematical teachers and students,
especially naval instructors and students of naval colleges.
0. H. DOWLINGr, C.E.
p
CONTENTS.
Pag©
Definitions
1
Polar Triangle
2
Fundamental Formula
4
Relations between the Sides and Angles of Spherical Triangles . . 5, 7
Napier's Analogies
8
Right-angled Triangles
10
Napier's Circular Parts
11
Solution of Oblique-angled Spherical Triangles
18
Ambiguous Cases of Spherical Triangles . . . ' 23
Table of Results from the Ambiguous Cases
25
To reduce an Angle to the Horizon
26
Numerical Solution of Right-angled Spherical Triangles
28
Quadrantal Triangles
38
Oblique-angled Triangles
35
Area of a Spherical Triangle .
44, 48
Girard's Theorem
51
Legendre's Theorem
52, 55
Solidity of a Parallelepiped
55
LexelFs Theorem
57, 59
Polyhedrons
61, 67
Examples and Problems
67
SPHEEIGS.
r.
PRELIMINARY CHAPTER.
]. A sphere is a solid determined by a surface of which all
the points are equally distant from an interior point, which is
called the centre of the sphere.
2. Every section of a sphere
made by a plane cutting it, is the
arc of a circle.
c——
Let 0 be the centre of the
sphere, A P B A a section made by
1
a plane passing through it, draw
00 to the cutting plane, and pro¬
N
M
duce it both ways to D and E, and V
draw the radii of the sphere 0 A,
OP. '
Now, since OOP and OCA are
right anales,
O A2 2 — 0 C22 = A C2. 2
2
and OP -OC = PC , but 0A = OP2;
AC2 = PC2
or AC = PC; hence the section APBA is a circle.
If the cutting plane pass through the centre, the radius of
the section is evidently equal to the radius of the sphere, and
such a section is called a great circle of the sphere.
3. The poles of any circle are the two extremities of that
diameter or axis of the sphere which is perpendicular to the
plane of that circle ; and therefore either pole of any circle is
equidistant from every part of its circumference, and, if it be
a great circle, its pole is 90° from the circumference. A
spherical triangle is the portion of space comprised between
three arcs of intersecting great circles.
4. The angles of a spherical triangle are those on the surface
of the sphere contained by the arcs of the great circles which
form the sides, and are the same as the inclinations of the
planes of those great circles to one another.
5. Any two sides of a spherical triangle are greater than the
third side.
B
9
PPHERIC8.
Since by Euclid XI. 20, any two of the plane angles, which
form the solid angle at (), are together greater than the third,
hence any two of the arcs which measure those angles must
be greater than the third.
6. Since the solid angle at 0 (see fig. p. 3) is contained by
three plane angles, and by Euclid XL 21, these are together
less than four right angles, hence the three arcs of the sphe¬
rical triangle which measure those angles must be together less
than the circumference of a great circle, that is a-{-S + c>
360, and since any two sides of a triangle is greater than the
third, we have 6 + c>a; a b.
ON THE POLAR OR SUPPLEMENTAL TRIANGLE.
7. If three arcs of great circles be described from the angular
points A, B, C, of any spherical triangle A B 0, as poles, the
sides and angles of the new triangle, D F E, so formed will
be the supplements of the opposite angles and sides of the
other, and vice versa.
Since B is the pole of D F, then B D is a quadrant, and
since C is the pole of D E,f C I) is a quadrant; therefore the
distances of the points B and C from D being each a quadrant,
they are equal to each other, hence D is the pole of B C.
n DE = 180°—C; EF = 180o-A;
AFD = 180°—B; and D = 180o-BC;
E == 180°—AC; F == 180°—AB.
, Also, AB=180o—F; BC = 180o—D;
A
AC = 180°—E; A = 180°- FE; .
r-^E B = 180°—FD; C = 180°—DE.
The sum of the three angles of a spherical triangle is
greater than two right
angles, and less than six right angles.
For if a' -f b' + c' be
the sides of the supplemental or
polar triangle, A = ISO0—a'; B = 180°—6'; C == 180°—'c';
hence A + B-fC + a'+t'+c' = 6 x 90 = G right angles;
but a'+ &'+ c is less than four right angles, by Euclid XL;,
21; therefore A + B + C is greater than two right angles; and
as the sides a', b', c', of the polar triangle must have some
magnitude, the sum of the three angles A, B, C must be less
than six right angles.
SPHERICAL TRIGONOMETRY.
CHAPTER I.
8. Spherical Trigonometry treats of the various relations
between the sines, tangents, &c., of the known parts of a sphe¬
rical triangle, and those that are unknown; or, which is the
same thing, it gives the relations between the parts of a solid
angle formed by the inclination of three
planes which meet in a point, for the solid ?.
angle is composed of six parts, the incli- ;
c
nations of the three plane faces to each \ \ jj/1 "v"-other, and also the inclinations of the ' X/
three edges; in fact, a work might be writ- \ /
ten on this subject without using the /
spherical triangle at all, for the six parts \ \ /
of the spherical triangle are measures of the
six parts of the solid angle at 0. See fig.
9. If a spherical angle have one of its angles a right angle,
it is called, a^ right-angled triangle; if one of its sides be a
quadrant, it is called a quadrantal triangle ; if two of the sides
be equal, it is called an isosceles triangle, &c., as in Plane
Trigonometry.
10. To determine the sines and cosines of a spherical tri¬
angle in terms of the sines and cosines of the sides.
Let 0 be the centre of the sphere on which the triangle
A B 0 is situated, draw the radii OA, OB, 0 0; from 0 A
draw the perpendiculars A D and A E, the one in the plane
0 AB, and the other in the plane 0 AC, and suppose them to
b2
3:
SPHEIUCAL TRIGONOMETRY.
meet the radii OB and 0 0 produced in D and E. The angle
D A E is equal to the angle A of the spherical triangle, and
taking the radius unity we have A D = tan C, 0 D = sec C,
A E = tan b, 0 E = sec b.
Then in triangles DAE and D 0 E we have
O D2 + 0 E2 - 2 0 D . 0 E cos E 0 D = D EJ
AD2 + AE2 —2AD . AE cos A = DE2
by subtracting
the second equation from the first, observing
that OD2— AD2 = OE2— AE2= 1, and EOD is measured
by B C or a, we obtain
2 + 2 A D . A E cos A — 2 0 D . 0 E cos a — 0;
or by substituting the above values
1 + tan b . tan c cos A — sec b sec c cos a = 0
,
1 ,
sin b
but sec b7 j-, tan o = :
cos 6
cos b
1
sin c
sec c
, tan c = :
cos c cos c
sin b sin c cos A
cos a
1 H
7
=
cos
b cos c
cos bj cos
c 0;
hence cos a = cos b cos c + sin b sin c cos A
(1)
which is the fundamental formula in Spherical Trigonometry.
11. In the figure the sides b and c are less than 90°, but it is
easily seen that equation (1) is gene¬
ral. Let us suppose that one of the
sides, A C or b for example, is greater
than 90°; draw the semi-circumj.b ferences C A 0', 0 B C, and make the
...
- triangle ABC of which the sidea
v
a' and b', or B C and A C, are supc7
•*'" plements of a and b, and the angle:
B A C the supplement of A. Since
the sides b and c are less than 90°, the equation (1) can be
applied to the triangle A B C, and gives
cos a' = cos b' cos c -f- sin sin c cos B A C ... (2).
SrHERICAIi TRIGONOMETRY.
Now a! = 180°- o, V = 180°-b, B A C = 180°- A; these
values substituted in eq. (2) will give eq. (1), which shows
that it is true for the case where b is greater than 90°.
Let us now suppose that the two
sides b and c are both greater than
90°; produce AB and A C till they b, / ;
intersect in A', which forms the c ./ /c'
triangle B C A' in which the angle A'
/
is equal to A, and the sides b' and c'
/B
the supplements of b and c; by
making the substitutions in this
case, we still find that equation (1)
satisfied.
Lastly, we can verify equation (1) in the case where
I = 90° and c = 90° either both together or separately.
If we apply equation (1) to each of the sides of the triangle,
we shall have three equations by means of which we can
always find any three parts whatever of the triangle, when the
three others are given. But, for practice, it is necessary to
have separately the divers relations which exist between four
parts of the triangle taken in every possible manner. There
are in all four distinct combinations, which we proceed to
give.
12. 1st, Relation between the three sides and an
By applying equation (1) to the three angles, we
cos a = cos b cos c + sin b sin c cos A
cos b — cos a cos c + sin a sin c cos B
cos c = cos a cos b -f sin a sin b cos C
angle.
have
(1)
(2)
(3)
13. 2nd, Relation between two sides and their opposite
angles.
From equation (1) we have
.
cos a — cos b cos c
cos A = ;—7—;—
sin o sin c
tt
. 2. A
a —« ,cos. b»
cos ~
c)s
A = -1 — cos22A
Hence
sin
A = .1 — (cos .
sin'4 o sm-'c
(1 — cos2 6) (1 — cos2 c) — (cos a — cos b cos c)3
sin2 b sin2 c
6
.
SPHERICAL TRIGONOMETRY.
sin A
yi — cos2 a — cos2 b — cos2 c + 2 cos a cos b cos c
sin a sin a sin 6 sin c
We must take the radical with the positive sign, seeing
that the angles and the sides are less than 180°; their sines
are positive. As the second member remains constant when
we change A and a into B and b, &c., we have
sin A
sin a
sin B
sin b
sin 0
sin c
Hence in any spherical triangle, the sines of the angles
are to each other as the sines of their opposite sides.
14. 3rd, Relation between the two sides and their included
angle, and the angle opposite one of them.
In considering the combination a, A, 0; first eliminate
cos c, between the equations (1) and (3) and we have
cos a = cos a cos2 b + cos b sin a sin b cos 0 + sin b sin c cos A
transposing cos a cos2 b, and observing that cos a—cos a cos2 b
= cos a sin2 b; and, dividing the whole by sin b sin a, it
becomes
cos a sin b ,
^ , sin c cos A
; = cos b cos 0 H ; ,
sm a
sin a
sin c
sin 0
,
.
,
, .
but —r-; and consequently we have for the relation
sin a
sin A
sought
cot a sin b = cos b cos 0 + sin 0 cot A.
By permuting the letters, we have in all the following six
equations:
cot a sin b = cos b cos 0 + sin 0 cot A
cot b sin a = cos a cos 0 + sin 0 cot B
cot a sin c = cos c cos B + sin B cot A
cot c sin a = cos a cos B + sin B cot 0
cot b sin c = cos c cos A + sin A cot B
cot c sin b = cos b cos A 4- sin A cot 0
(5)
(6)
(7)
(8)
(9)
(10)
SPHERICAL TRIGONOMETRY.
7
15. 4:th, Relation between one of the sides and the three
angles. Eliminate b and c from the equations (1) (2) (3) : to
do this we have by the last article
cos a sin & _
^ sin c cos A
; = cos o cos (J
— .
,
sin a sin a
sin b sin B
_ sin c sin 0
and since ——=-—- and ——=-—
sin a smA
sin a sin A
we have
cos a sin B = cos b sin A cos 0 + cos A sin C;
and changing a and A into b and B, and vice versa, we obtain
cos b sin A = cos a sin B cos 0 + cos B sin C.
We have only now to eliminate cos b by the two preceding
equations. We find after reduction the relation sought
between ABO and a, which, applied to the three angles
successively, will give the three equations
cos A = — cos B cos 0 + sin B sin 0 cos a
(11)
cos B = — cos A cos 0 + sin A sin 0 cos b
(12)
cos 0 = — cos A cos B + sin A sin B cos c
(13)
16. The analogy of these equations with the fundamental
formula is striking, and conducts us to a remarkable conse¬
quence. Let us imagine a spherical triangle A' B' 0', of which
the sides a' V c' are the supplements of the angles A, B, C;
then from equation (1) we shall have
cos a' — cos V cos c' + sin b' sin b1 cos A'.
Now sin a'= am A, cos a = — A, sin U = sin B, &c., then
— cos A = cos B cos 0 + sin B sin 0 cos A!.
From this equation we find for cos A! a value equal but of
a contrary sign to that which we find for cos a in equation
(11); then a = 180° — A', similarly b — 180 — B', and c =
180 — C. Hence, having given any spherical triangle, if
we describe another triangle,, the sides of which are the sup¬
plements of the angles of the first, then the sides of the first
will be the supplements of the angles of the second. From
this property the two triangles are called supplementary, and
eometimes the triangles are said to be polar to each other.
8
SPHERICAL TRIGONOMETRY.
NAPIER S ANALOGIES.
17. We now proceed to deduce the formulse known by the
name of the analogies of Napier, which are employed to sim¬
plify some of the cases of spherical triangles.
The equations (1) and (2) give
cos a — cos b cos c = sin b sin c cos A;
cos b — cos a cos c = sin a sin c cos B.
By division, observing that S^n a
^^
sin b
sin B '
cos
b
—
cos
a
cos
c
sin
A
cos B
we have
cos a — cos b cos c
sin B cos A *
By subtracting and adding unity to both sides of this
equation and again dividing
cos b — cos a 1 -f cos c
sin (A — B)
cos b + cos a 1 — cos c
sin (A + B)"
But by Plane Trigonometry, page 30, (12),
cos b — cos a
= tan (a 4- b) tan J (a — b)
cos b + cos a
but1 + 003 c
1 — cos c
tan2 ^ c
and sin (A + B) = 2 sin % (A + B) cos J- (A + B)
sin (A — B) = 2 sin ± (A — B) cos J (A — B)
Substituting these values, the above equation becomes
tan J (a + b) tan (a — b) =
tan2 i ccos j(A-B)\_
\8in J (A + B) cos | (A + B) /
, .
sin a sin A
ana since-:—-—sin b sin B
(a)
sin a
h
+
_ sin A + sin B
sin a — sin b sin A — sin B*
By Plane Trigonometry, page 30,
tan | (q + b) sin | (A + B) cos | (A — B)
tan j (a — b)
cos £ (A + B) sin | (A — B)'
we have
SPHERICAL TRIGONOMETRY. 9
Multiply th^se two equations together, and then dividing
one by the other and extracting the root, observing that
tan £ (a + 5) and cos J (A + B) ought to have the same
sign,
t™
+
^~g -
(H)
<15)
To apply these to the polar triangle we must replace a, b, c,
A,B,0, by 180°-A, 180o-B, 180o-C, 180o-.a, 180o-b,
and there results,
+
=
(1C)
=
(17)
My able and talented friend, Mr. Eeynolds, of Chelsea Hos¬
pital, has sent me the following very neat method of deducing
Napier's Analogies, which he says was communicated to him
by Mr. Adams, the celebrated astronomer of Cambridge.
sin A
sin B
sin A + sin B
T ,
Let
m=
= = — ;——.
sin a
sm b
sm a + sin b
Then by the formula (11) page 7,
cos A + cos B cos C = sin B sin C cos a
= m sin C sin b cos a...(I)
cos B + cos A cos 0 = sin A sin 0 cos b
= m sin C sin a cos b...(2)
Add (1) and (2), then
{cos A + cos B} (1 + cos 0) = m sin 0 sin (o + b).
Also sin A + sin B = m (sin a + sin b)..
Dividing and reducing we have
a—b
10
SPHERICAL TRIGONOMETRY.
Again, subtracting (2) from (1)
(coa B — cos A) (I — cos 0) = vi . sin (a — b) sin 0
and sin B + sin A = m (sin a + sin 6);
dividing
a—b
sm
j^q
2
tan . tan _ = —— (Anal. 2)
8m
—2—
The other two follow of course from the polar triangle.
ON RIGHT-ANGLED SPHERICAL TRIANGLES.
18. The preceding formulae will apply to right-angled
triangles, if we make any one of the angles = 90°.
If A = 90° we have
cos a = cos b cos c
(1)
sin b = sin a sin B
(2) sin c = sin a sin 0 (7)
tanb = tan a cos C
(3) tanc = tana cos B..;... (8)
tan b = sin c tan B
(4) tan c = sin b tan 0 (9)
cos B = sin 0 cos b
(5) cos C^= sin B cos c
(10)
cos a = cot B cot 0 (6)
These six independent formulae are all adapted to logarith¬
mic calculation.
The first gives a relation between the hypothenuse and the
two sides containing the right angle; the second, one side
and angle opposite; the third, between the hypothenuse, a
side, and the adjacent angle; the fourth, between the two
sides and the angle opposite to one of them; the fifth, between
one side and the two oblique angles; lastly, the sixth, between
the hypothenuse and the oblique angles.
19. The formula (1) requires that cos a must have the same
sign as the product cos b cos c, or that the three cosines must
be positive, or that only one must be so. Therefore in any
right-angled spherical triangle the tnree sides must be less
than 90°; or two of them must be greater than 90°, and the
third less. The formula (4) shows that tan b has the same
sign as tan B, and tan c the same sign as tan 0. Therefore
each side containing the right angle is of the same kind or
affection as the angle opposite, that is, the angle and the side
are both less than 90° or both greater.
SPHERICAL ^TRIGONOMETRY.
11
napier's circular parts.
20. As we have before observed, the above formulas are
simple and well adapted for logarithmic computation, yet
they are not easily remembered; therefore it is of importance
that we should have some method which will relieve the
memory as much as possible; this is supplied by what is
termed'Napier's Circular Parts. By committing to memory
the two rules which will be given hereafter, the student will
he able to solve all the cases in right-angled triangles, as well
as if he had all the formulas by heart.
The circular parts of a right-angled spherical triangle are
five, namely, the two sides, the complement of the hypothenuse, and the complements of the two angles (the right angle
being always omitted).
Three of these circular parts, besides the right angle, enter
every proportion, two of which are given, and the third
sought.
These three parts are named from their positions with
respect to one another, that is, according as they are joined
or disjoined, observing that the right angle does not separate
the sides.
If the three circular parts join, that which is in the middle'
is called the middle part, and the other two are called extremes corijunct.
If the three circular parts do not join, two out of the five
must, and that part which is separate or alone is the middle
part, and the other two are called extremes, disjunct.*
These things being understood, the following is the general
rule.
The sine of the middle part is equal to the product of the
tangents of the extremes conjunct.
* Thus, if in figure page 12, we suppose B C, the angle B, and the
fide A B to be the quantities that are to be used; now as they lie altoge¬
ther, the angle B is the midale part, and the two sides, B C and A B, are the
extremes conjunct. Also, if the angle B, A B and A C be the quantities,
then since the right angle does not separate the sides, A B is the middle part
and the other two elements are the extremes conjunct. But if the quan¬
tities be A C, B C and the angle B, then the angle C is said to separate A C
from B C, and the'side A B is said to separate A G from the angle B, that
part A C which is separated from both the others, call the middle part, and the
parts which are disjoined from it call extremes disjunct. This practical
method will be useful to seamen, and requires very little effort of memory.
12
SPHERICAL TRIGONOMETRY.
The sine of the middle part is equal to the product of the
cosines of the extremes disjunct.
From these two equations, proportions may be formed,
observing always to take the complements of the angles and
hypothenuse; and that the cosine of a complement is a sine,
and the tangent of a complement is a co-tangent, and vice
versa*
21. Case 1. When the hypothenuse B C
c,
and the base A B are given to find the
/a*
remaining parts of the triangle.
Let us proceed to find A 0.
c
Here the hypothenuse and the two
sides are the three circular parts.
The hypothenuse being separated or disjoined from the
sides it is the middle part, and the sides are the extremes
disjunct.
Then sin B 0 = cos A B cos A G.
And since we must always take the complements of the
hypothenuse and angles, this becomes
cos a = cos h cos c.
Now, as this agrees with equation (1), the rule is proved
in this case.
To find the angle B.
Here the three circular parts all lie together, taking B
to be the middle part, then A B and B 0 are adjacent parts,
or extremes conjunct.
.v sin B = tan B 0 . tan A B;
taking the complements of B and B 0, we have,
cos B = cot a tan c,
which corresponds with (8), and therefore the rule is proved
in this case also.
To find the angle 0.
Here the side A B is separated from the hypothenuse by
the angle B, and it is separated from the angle 0 by the side
A 0, then A B being the middle part, the hypothenuse and
the required angle are the extreme or disjoined parts.
sin A B = cos B 0 cos 0;
taking the complements of B 0 and 0,
sin c =.sin o sin 0.
This agrees with (7), and therefore proves the rule.
SPHERICAL TRIGONOMETRY
13
22. Case 2. Given the two sides &, and c, which include the
right angle, to find the hypothenuse and the angles.
1. To find the hypothenuse.
As the two sides are separated from the hypothenuse they
will be extremes disjoined, the hypothenuse being the middle
part;
sin B 0 = cos A B cos A C;
taking the complement of B 0,
cos a = cos b cos c; which is the same as equation (1);
a
cos c == cos
=.
cos 6
To find angle 0.
Since the right angle A does not disjoin, the three parts all
lie together, hence A 0 being the middle part, A B and angle
0 are the adjacent parts, or extremes conjunct.
sin A 0 = tan A B . tan 0 ;
taking the complement of C,
sin b
tan c cot c; which agrees with (9);
, /n
tan c
or tan
U = ——
Bin 6
To find the angle B.
The three circular parts all lie together again, A B being
in the middle ; calling it the middle part, then A C and angle
B will be extremes conjunct.
sin A B = tan A 0 tan B;
taking the complement of B,
sin c = tan b cot B, which agrees with (4);
, -o
or tan
B = tan b
sm c
23. Case 3. Given the hypothenuse a and angle B to find
c, G.
1. To find AO or b.
As A 0 is separated from the hypothenuse by the angle 0,
and from the angle B by the side A B; calling A 0 the middle
part, then B 0 and angle B are extremes disjunct.
sin A 0 = cos B 0 cos B;
taking the complements of BO and 0,
sin b =sin a sin B, which is the same as equation (2).
SPHERICAL TRIGONOMETRY*
2. To find A B or c.
Here the three circular parts all lie together, and AB in the'
middle ; calling it the middle part, then A B- and B C will be
adjacent parts^ or extremes conjunct.
sin B = tan AB tan B 0 ;
taking the complements of B and B 0.
cos B = tan c cot a;
tanc=
C0S
= cos B tan a. which agrees with equ. (8).
cot a
-aw
3. To find G.
Here the circular parts lie all together, the hypothenuse
being in the middle ; call it the middle part, and the angles
B and 0 will be adjacent parts.
sin B 0 = tan B tan G;
taking the complements throughout,
cos a = cot B cot C ;■
cot C =
= cos a
tan
B, which agrees with equ. (6).
24:. Case 4^ Given the side A C or b and the opposite anglQ
B to find a, c, 0.
1. To find the hypothenuse B 0 or a.
Here ^ or AO is separated from the hypothenuse by the
angle 0, and from the angle B by the side AB; calling then
A 0 the middle part, the angle B and the hypothenuse are
the extremes disjunct.
sin A 0 = cos B C cos B;
taking the complements of B 0 and B,
sin b ==sin a sin B which agrees with equation (2).
sin a = sin
. &.
smB
2. To find c.
Aa the right angle does not disjoin, c lies in the middle
between b and B; calling it the middle part, b and B are the
extremes conjunct.
sin 0 = tan b tan B;
taking the complement of B,
eih c = tan b cot B, which agrees with equation (4).
SPHERICAL TRIGONOMETRY»
15
3. To find C.
Here the angle B is separated from c by the hypothenuse,
and it is separated from b by the side A B; calling it the
middle part,
sin B = cos b cos C ;
taking the complements of B and 0,
cos B = cos b sin 0, which agrees with equation (5).
cos B
sin 0 :
cos b
There is here an ambiguity, since each quantity is deter¬
mined by its sine, and we see that this really ought to be the
case. In fact, if the triangle BAG (fig. p. 12) right-angled at
A, satisfy the equation; produce B A and B 0 till they inter¬
sect in D, then take D A' = B A, and D C7 = B 0, the triangles
B A 0, D A' O7 will be equal in all respects, then the angle A is
a right angle, and G' A! = OA = b. Thus the triangle B A'C
is right-angled, and contains also the given parts B and b;
we can therefore take at will a < 90°, or A > 90°, but when
the choice is once made the affection or species of c will be
determined by the equation cos a = cos b cos c, and that affec¬
tion will be the same as that of C. There will be only one
triangle which has two right angles when b = B, and none
when we have sin b > sin B.
25. Case 5. Given the side b and the adjacent angle C,
to find a, c, B.
1. To find the hypothenuse a.
Here the parts all lie together, the angle 0 being in the
middle; call it the middle part, then A G and B 0 are the
extremes conjunct.
sin 0 = tan A 0 tan B C;
taking the complements of 0 and B 0,
cos C = tan b cot a; which agrees with equation (3);
.a = tan
b
. . tan
—.
cos 0
2. To find AB or c.
As the right angle does not disconnect, A 0 is the middle
part, and A B and angle 0 are the extremes conjunct,
sin A C = tan A B tan 0;
16
SPHERICAL TRIGONOMETRY.
taking the complement of 0,
sin b = tan c cot c;
tanc =^= S^Q ^ == sin & tan C, which agrees with equ. (9)r
cotO
3. To find the angle B.
Since the angle B is separated from b by the side A B, and
from the angle 0 by the hypothenuse BO; calling it the
middle part, then A 0 and the angle 0 are the extremes
disjunct.
sin B = cos AO cos 0;
taking the complements of B and 0,
cos B = cos b sin 0, which agrees with equation (5).
Here a, c and B are found without any ambiguity.
26. Oase 6. Given the two oblique angles B and 0 to find
a, b,c.
1. To find a or B 0.
Here a, B and 0 all lie together, a or B 0 being in the
middle; call it the middle part, then B and C are the ex-'
tremes conjunct^
sin B 0 = tan B tan 0;
taking the complements of the whole,
cos a = cot B cot 0, which agrees with equation (6).
2. To find & or A 0.
Here Bis separated from0by the hypothenuse BO, and it
is separated from A 0 by the side A B; calling it the middle
part, then the angle 0 and A 0 are the extremes disjunct.
sin B == cos 0 cos A 0 ;
taking the complements of B and 0,
cos B = sin 0 cos &, which agrees with equation (5).
3. To find c or A B.
Here the angle 0 is separated from A B by the side A 0,
and from the angle B by the hypothenuse B 0; calling it the
middle part, then A B and the angle B are the extremes
disjunct.
sin 0 = cos A B cos B;
taking the complements of 0 and B,
cos 0 = cos c sin B, which agrees with equation (10);
SPHERICAL TKIGONOMKTRY.
17
cos 0
These values leave no ambiguity, and if the triangle is
impossible they will show that it is so.
27. When a'triangle is isosceles, the two equal sides are
only counted as one element, and the angles which are oppo¬
site to them also as only one element. Now, if we draw the
arc of a great circle through the vertex of the triangle and
the middle of the base, we divide it into two right-angled
triangles, equal in all respects, and in each of which we
know two elements besides the right angle, then the isosceles
triangle can be solved by the i'ormulai for right-angled
triangles.
28. If in a spherical triangle ABC, in which we have a + 6 =
180°, produce a and c till they intersect in D, we shall have
a + CD = 180°, hence CD = 5; therefore,
the solution of the triangle A B C is brought c
to that of the isosceles triangle A B D.
The same thing may be said of a
triangle, in which two angles are the a <-•
spherical supplements of each other, for
we cannot have A -f- B=180o without at the same time having
A + B —180° and vice versd. In fact, in the isosceles triangle
A CD, the angle CAD=D = B. Now, CAD + CAB =
ISO0; then also,
in the triangle A B C we ought to have
A + B = 180o.
18
SPHEKICAL TKIGONOMETRV.
CHAPTER II.
SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES.
29. Case 1. Given the three sides a> b, c to find the angles
A, B, 0.
To find A, we have by equation (1)
.
cos a — cos b cos c
COS A =
; = ; >
sin o sin c
but we obtain an expression better adapted to logarithms by
finding sin J A, cos % A, &c., as in Plane Trigonometry.
Since 2 sin2 i A = 1 — cos A, we have by substituting the
above value of cos A,
_ . - .
^
cos a — cos b cos c
2 sm20 i A
1 :— —;
^ sin u7 sin c
cos b cos c + sin b sin c — cos a
sin b sin c
cos (b — c) —cos a
sin b sin c
(by equation (8) page 30, Plane Trigonometry,)
2 sin ^ (a + 6- c) sin -J- (a — b + c)
5
sin b sin c
sin IL A = . /Bi'1 i (« + b — c) sin j (a — 6 + c)
V sin b sin c
For the sake of abridgment, put a + 6 + c=:2s, and the
preceding expression becomes
/a (
n(
c)
»A=v
'T\o)°
V "sin
sin c'- .
me way
In the same
/Bi'iSBin (s-q).
cos iA== V
sin b sin c
_ .
/sin (s— b) sin (5 — cV
. tan i A = A / — > —v v !
V sin 5 sm (s — a)
SPHERICAL TRIGONOMETRY. * 19
30. Case 2. Given the two sides a, 5, and the angle A op¬
posite to one of them, to find c, B, 0.
We obtain at first the angle B opposite to b by the
proportion
sin a : sin b : : sin A : sin B ;
. sm
. o7 =
sin A sin .
b
sin a
It will be best to determine c and 0 by Napier's Analogies,
which give
tan i c = tan i2 V(a — b); Bin^(A + E)
^
ein£(A — B)
6in
f (« + &),
sin i (a — b)
As the angle B is determined by its sine, it can either be
acute or obtuse. However, for certain values of the given
quantities a, b, A, there will be only one triangle. We may
refer back to the similar case of plane triangles; we can thus
find 0 in a direct manner by the equation
cot A sin 0 + cos b cos 0 = cot a sin b.
To effect this, let us at first determine an auxiliary angle 0
by putting cot A = cos b cot 0, from whence we have
cot i 0 = tan i (A — B) .
eot6=e^A;
cos b
then in the equation (5), p. 6, cot A=cos b cot 0=
cos
? CQS ^
sin $
the equation becomes
cos b (sin 0 cos <j) + cos 0 sin $) = cot a sin b sin 0
from which we find
sin (0
v +
r/
tan a
hence 0 + 0 is determined; let 0 + 0 = m, and we have
C = m — 0.
After having found 0, we obtain the side c by the proportion
sin A : sin 0 : : sin a : sin c.
But if we wish to find c directly, we must refer back to
equation (1), page 5,
cos b cos c + cos A sin b sin c = cos a.
20
SPHERICAL TRIGONOMETRY.
This may be reduced in the same way as the equation
above, by using an auxiliary angle 0, putting cos A sin b =
cos b cot 0, from whence we have
cot 0 = cos A tan b;
consequently, the above equation becomes
cos b (sin 0 cos c + cos 0 sin c) = cos a sin 0, or
• (c
/ +a 0)
/\ =cos cl sm 0
sin
cos 6
Having found 0, we can easily find c.
31. Case 3. Given the two sides a and b and the included
angle 0 to find A, B, c.
The formula (5) (6), page 6, give for A and B
cot A =.cot a sin & — cos b cos 0
sin 0
cot
b
sin
a
— cos a cos 0
cot B =
siii 0
By employing auxiliary angles it is easy to reduce each
numerator to a single quantity, but it is more simple to recur
to Napier's Analogies.
tan I (A + B) = cot * 0 .
008
f ~
cos f (a + o)
tan J (A - B) = cot £ 0 . S|n f ~ ^
sin & (a + b)
which give J (A + B) and -J- (A — B), consequently by adding
and subtracting we find A and B.
The angles being found we obtain c from the proportion
sin A : sin 0 :: sin a : sin c; but if we wish to have c directly
we must take the formula, page 5,
cos c = cos a cos b + sin a sin b cos 0;
in which, if we make sin b cos C = C08 ^ CQ3 0
sin0
= cos b cot 0, then it becomes without any ambiguity
co
^
n
s & sin (a + 0\
cot 0 = tan b cos 0, .. cos c=
;—^—' w.
sm 0
32. Oase 4. Given the two angles A and B, and the adja¬
cent side c, to find a, b, c.
SPHERICAL TRIGONOMETRY. 21
We can find a and b by the formulae (7) and (9), page 6,
cot A sin B ^+ cos—
B cos c
cot. a =
sin c
cot. 0t — cos B sin A r-i+ cos A cos c
sin c
and better still by Napier's Analogies,
tan i (a
+ b) = tan J c . CQ3 ^ ^ ^
- v ^ ^
cos i (A + B)
tan J2 V
(a — b) = tan J2 c . S^n f
~
sin i (A + B)
These equations determine J (a -f- b) and J (a — b), and
from which, by adding and subtracting, we find a and b.
We can now find 0 by the proportion
sin a : sin c :: sin A : sin 0,
or we can find 0 directly by making use of the formula, equa¬
tion (13), page 7, viz.,
cos 0 = sin A sin B cos c — cos A cos B.
If we put sin B cos c = cos B cot0, it will become
^ x B
-n
n
cosB sin
(A —-jp)
cot0=tan
cos c, cos
0=
^—3
—/.
sm <p
This case is analogous to the third case, and offers ro
ambiguity.
33. Case 5. Given the two angles A and B, and the side a
opposite to one of them, to find 6, c; 0.
This case is quite analogous to the second, ai^d is treated
in the same manner, and has the same ambiguities.
We deduce b from the proportion
sin A : sin B :: sin a : sin b,
and we find c and 0 by the formulae already employed,
The side c can also be obtained by equation (7),
cot a sin c — cos B cos c = cot A sin B,
in which we make cot a = cos B cot 0
22
SPHERICAL TRIGONOMETRY.
. „ . .
cot a . ,
,v
tan B sin 0
. . cot0 =
sin (c.— 0) = — t *
cosB tan A
Lastly, we can find C, for sin a : sin c :: sin A : sin 0, or
better by means of the equation,
cos a sin B sin 0 — cos B cos G = cos A,
we reduce the first member to a monomial by putting
cos a sin B = cos B cot 0, from whence we have
icos a tan
cos
A v»—
sin 0 >
cot, 0=
B, sin (0
v — 0) =
'
cos B
these values determine 0, 0—0,and consequently the angle 0.
34:. Case 6. Given the three angles A, B, 0, to find the
sides a, b, c.
This case is solved in a similar way to the first.
By page 7, equation (11)
cos a + cos B cos C
cos a
= sin -p
. n
B sm
C >
and by the same method, as used in the first case,
sin | a^ /si'iSsin^A-S)
V
sin B sin U
• cos i a =
/sin (B - 8). ein (G--S)
V sin B sm 0
sin S sin (A
— S)7 >.-'
tan J a = a // -^-7^
gx \ /n
Q
V sin (B — S) sm (0 — o)
By using the polar triangle in Case 1, we have
. 1
/ — cos S . cos (S — A)
sm a = a /
. t> • n
V sin B sm 0
/cos (S-B) cos (S —0)
COS -hJ a = x /
^ ; =r^ ^
V sm B sm 0
- /— cos S . cos (S — A)
tan & a ios (S _ B) cos (SC)
A
The first and last of these appear under an impossible
form, but since S is always greater than 90 and less than 270,
the cos S is alwavs negative, and therciore makes the quan¬
tity under the radical always positive.
SPHERICAL TRIdONOMETRY.
23
ON THE AMBIGUOUS CASES OF SPHERICAL TRIANGLES.
35. The only cases in which there is any uncertainty are the
second and fifth. We proceed to show in this article what
conditions are necessary that there may be two solutions, or
only one, or even when the triangle is impossible.
Let us consider upon a sphere a semicircle D 0 D' perpen¬
dicular to a whole circle 1) H D'; take
G D less than 90° and draw the arcs of
great circles OB, CB', C H. . . . from
the point 0 to the different points of
the circumference D H D'. Produce 0 D,
making 0' D = 0 D, and join 0' B. The
triangles C D B, CD B have a right angle
contained between the equal sides,
therefore C B = C B. Now we have
0 D C < 0 B + B 0', therefore 0 D < C B.
Hence, in the first place, the arc C D is the least that we
can draw from the point C to the circumference D H D'; and
consequently 0 D' is the greatest.
LetDB'=DB; then in the two triangles CD B and CDB'
have the two sides C D, C B and the right angle C D B of the
one, equal to the two sides C D, D B, and the right angle
C D B' of the other, hence C B' = 0 B. Therefore, in the
second place, the oblique arcs equally distant from C D or
C D' are equal.
Lastly, let D H > D B; draw C H and produce C B till it
intersects C H in 1. Then, since the arc 0 C is less than a
semicircle, it will meet 0 B produced beyond the point C;
this requires that the intersection I falls between H and O.
We have therefore C B < CI + I B, and consequently O B
+ B C < C I + 10. But we have I 0 < IH + H C, and
therefore 0'I + IC<0'H+H0; hence, fortiori, C B
+ B 0 < C H + H C. Now, C'B ==B 0 and C'H = H 0,
therefore we.have B 0 <C H C. Consequently, in the third
place, the oblique arcs are greater the farther they are from
0 D, or the more they approach 0 D'.
Now, suppose we have to construct a spherical triangle, the
given quantities being o, 6, and the angle A opposite to a.
We may at first remark that certain cases of impossibility
are indicated even by the calculation. To show this, make
24 . SPHERICAL TRIGONOMETUY.
the angle CAB =■ A and A C = b, produce A C and A B
till they intersect in E, then let fall the perpendicular 0 D
upon A E.
The arc C D ought to be of the same affection or species as
^ C is U
the angle A by Art. 19; then, when A is acute, C D is the
ghortest distance from the point 0 to the semi-circumference
A E, and it is the greatest when A is obtuse.
In the first hypothesis the triangle will be impossible if
we have a < C D, which gives sin a < sin C D, and in the
second it will be impossible if we have a > C D, which gives
again sin A > sin 0 D.
Now, in the right-angled sj)herical triangle A CD, we have
sin 0 D = sin b sin A;
then, in both hypotheses we shall have sin a < sin b sin A.
On the other hand, when we seek the angle B of the unknown
triangle ACB, we have
. B
-D =
sin b;
sin A
sin
;
sin a
then this value of sin B will be > 1, which is impossible.
If we have a = 0 D, there will be only one right-angled
triangle, A C D, which will be possible, and it is that which
again indicates the value of sin B, which becomes sin B=l.
It is understood that the angle A is not equal to 90°.
Let us now examine the different relations of magnitude
which the given quantities a, b, A can represent.
Let A < 90° and b < 90° (fig. p. 23). Since A and b
are < 90°, A D is also < 90° by Art. 19 ; then A D D E ;
if now we have besides a
b, it is clear that we can place
between C A and C D an arc 0 B = a, and that on the other
side, between 0 D and OE, we can put another C B' = C B =
a ; that is to say, there are two triangles ACB and ACB'
which have the same quantities given, viz., a b A.
When a — b, the triangle ACB disappears, and there re¬
mains only the triangle A C B'.
When a -f- b = 180, or when a + b > 180, the point B'
coincides with E, or passes beyond it, and then no triangle
can exist.
SPHERICAL TRIGONOMETRY.
We can discuss in the same manner the other ^hypotheses.
The results are all contained in the following table. The
sign ^ signifies equal to or greater than; and the sign
eignifies equal to or less than.
two solutions,
a<&
one solution,
b <90° {
a>6 0
no solution.
a + &>180
two solutions,
a + 5<180o
A<90°one solution,
b>90° J a + &>1800
no solution.
a%b
two solutions,
) = 90o|
no solution.
a>6
a + 6 > 180°
b <90° a + b < ^180°
a < Cb
A>90o-
&>90o
& = 90(
a
^Cb
a-\-b :i8oo
ja>5
two solutions,
one solution,
no solution.
two solutions,
one solution,
no solution.
two solutions,
no solution.
one solution,
no solution,
no solution.
one
solution,
A == 90° \
no
solution,
b >90°
I a + 6 <180o
no solution.
solutions ad infinitum.
a
=
90°
> = 90° |
a<[or >90° no solution.
b <90°
a^b
a<&
a+b^b
f
«<6
By the properties of thfc polar triangle, we can apply the
results to the fifth case, where A, B, a, are given, only taking
care to change a, b, A into A, B, a, the sign > into <, and
the sign < into j>.
When the given quantities fall in a case where we ought
to have only one solution, the calculation will still indicate
two. But to discern which ought to be taken, it is sufficient
to observe; that the greater angle must be opposite to the
greater side, and conversely.
See Lefebure De Fourcy's Trigonometry.
2G
SPHERICAL TRIGONOMETRY.
Suppose,Tor example, that we have given A=112°, a—102°,
b = 106°. In the preceding table, among the cases which
correspond to A > 90°, we consider that where b 90°, and
among these that where a<j^6. We may observe besides, that
a -j- b = 208°, therefore a + b > 180°, we conclude from the
table that there is only one solution, and since b is > a, the
angle B is greater than A, therefore B is obtuse.
TO REDUCE AN ANGLE TO THE HORIZON.
•
36. Let B A C be an angle in an inclined plane, and AD the
vertical passing through A. Draw the horizontal plane meet¬
ing the lines A B, A 0, A D, in E, F, G; the angle E G F is the
horizontal projection of the angle B A 0, or, in other words, it
is the angle BAG reduced to the horizon. It is this angle .
EFG that we have to calculate, supposing the angles BAG,
B A D, 0 A D, to have been determined by an instrument.
The geometrical construction is easy, for the line AG being
arbitrary, we shall have sufficient quantities given to con¬
struct at first the right-angled triangle, E A G
and FAG, then the triangle E A F, and, lastly,
//'d EGF. The calculation of the
angle E G F is equally easy. If we describe a
/ ft 1 sphere from the centre A with any radius, the
/ L lines AB, AO, AD, where they meet the
\ /y'
sphere, will determine a spherical triangle
j BOD, of which the sides are known by means
of the given angles, and of which the angle
BDC of the triangle is equal to the required angle EGF.
Then by the first case of oblique-angled spherical triangles,
page 18, we have •
sintA=A/ein(*-\)s.in-<l^;
* V sin 6 sin c '
where A = BAG; b = BAD; c = GAD; s = ^ (a + b -}-c).
Let a = 4:7° 45' 39", b = 69° 49' 19", c = 80° 17' 36". We
shall have 2 s=1970 52' 34", s = 98056' 17"; 8-6=29° 6'58";
3 - c = 180 38' 41".
27
8PHERI0AL TRIGONOMETKY.
log Bin (s — 5)
9*6871552
log sin (s — c)
9 •50474:12
comp. log sin b
0*0275078
comp. log sin c
0*0062623
2 log sin J A
19*2256665
log sin A
9*6128332
I A = 24° 12' 27"*9, or A = 48° 24' 56".
37. The following properties of spherical triangles we
shall premise before entering on the numerical solution of
triangles.
Any side of a spherical triangle is less than a semicircle,
and any angle is less than two right angles.
For the limit of any plane angle is two right angles, and
this is also the limit of any plane face of a solid angle.
The sum of the three angles is greater than two right
angles and less than six right angles.
If the three sides of a spherical triangle be equal, the three
angles will also be equal, and vice versd.
If the sum of any two sides of a spherical triangle be equal
to 180°, the sum of their opposite angles will also be equal to
180°, and vice versd.
If the three angles of a spherical triangle be all acute, all
right, or all obtuse, the three sides will be accordingly all
less than 90°, all equal to 90°, or all greater than 90°, and
vice versd.
The sum of any two sides is greater than the third side,
and their difference is less than the third side.
The sum of any two angles is greater than the supplement
of the third angle.
The sum of the three sides is less than the circumference
of a great circle.
If any two sides of a triangle be equal to each other, their
opposite angles will be equal, and vice versd.
Since
.
cos a — cos b cos c
cos A =* :—;—: ,
sin b sin c
#
cos b — cos a cos c
COS B = : ;
sin a sm c
o2
28
SPHERICAL TRIGONOMETRY.
.
.
cos a — cos a cos c
It b — a, these expressions are each = ; :
sin a sin c
cos A = cos B or A = B;
that is, the angles at the base of
an isosceles triangle are equal,
and the converse of this holds
also. From this it is easily shown
that the greater side of a sphe¬
rical triangle is opposite the
greater angle, for let ABO be
greater than CAB, and make the angle A B D equal to the
angle DAB;
DA = D B.
AC = AD + DC = DC4-r)B,butDC + DB>BC
A 0 > B 0.
OX THE NUMERICAL SOLUTION OF RIGHT-ANGLED SPHERICAL
TRIANGLES.
38. When the hypothenuse and one side are given.
Ex. 1. Given the hypothenuse B C = 63° 56' 7", and the
side A B = 40°, to find the remaining parts of the triangle.
To find the other side, A 0.
Here the hypothenuse and the two sides are the three
circular parts.
The hypothenuse being separated or disjoined from the
sides by the angles is therefore the middle part, and the sides
the extremes disjunct.
sin B C = cos A B cos A 0;
taking the complement of hypothenuse as directed by the
rule,
cos B 0 = cos A B cos A C;
log cos B 0 = log cos A B + log cos A C — 10
log cos A 0 = log cos B 0 — log cos A B + 10
= log cos,G30 5G' 7"— log cos 40°+10
= 9-64284G4: - 9-8842540 + 10
= 9,7585924;
AO = 54° 59' 59"'G.
SPHERICAL TRIGONOMETRY. 29
The side A 0 is acute, because-the hypothenuse and the
given side have the same affection.
To find the angle B.
This angle connects the hypothenuse and the given side,
and is therefore the middle part, and the other the extremes
conjunct.
sin B = tan A B . tan BO;
taking the complements of the angle and hypothenuse,
cos B = tan A B . cot B 0;
log cos B = log tan A B + log cot B C — 10
= log tan 40° + log cot 63° 56' 7" •
= 9*9238135 + 9-6894:258 -10
= 9-6132393;
/. B = 65° 46' 5".
The angle B is acute, as the hypothenuse and given side
are of the same affection.
To find the angle 0.
Here the side A B is separated from the hypothenuse by
the angle B, and it is separated from the angle ,0 by the side
A 0; take it to be the midcjle part, then B 0 and the angle
A are extremes disjunct.
sin A B = cos B 0 . cos 0;
taking the complements of hypothenuse and angle 0,
sin A B = sin B 0 sin 0;
log sin A B = log sin B 0.4- log sin 0 — 10,
log sin 0 = log sin A B — log sin B 0 +10
= log sin 40° — log sin 63° 56' 1" + 10
= 9*8080675 + 0*04:65794 by taking comp. log.
63° 56' 7";
= 9-8546469;
0 = 45° 41' 21".
The angle 0 is acute, the hypothenuse and given side being of the same affection.
30
SPHERICAL TRIGONOMETRY.
\
When the two sides are given.
Given the side AO = 52° 13', and the side A B =? 42° 17',
to find the remaining parts.
To find the angle B. (See fig. p. 12.)
As the right angle does not disjoin, A B is the middle
part, and the angle and side A 0 are extremes conjunct.
sin A B = tan B . tan A 0;
taking the complement of B,
sin A B == cot B tan A C ;
log cot B = log sin A B — log tan A 0 + 10
= 9-8278843 + 10 - 10-1105786
= 9*7173057, which is the log cot 62° 27';
/. B = 62° 27',
which is acute, like its opposite side.
To find the angle 0.
Here A 0 is the middle part, and the angle 0 and AB are
extremes conjjinct.
sin A 0 = tan A B tan 0;
• taking the complement of 0
sin A 0 = tan A B cot 0;
log cot 0 = log sin A 0 — log tan AB + 10
= 9-8978103 - 9*958754:2 + 10
= 9*9390561, which is the log cot of 49°.
The angle is acute like its opposite side.
To find the hypothenuse B 0.
The hypothenuse being separated from the sides by the
angles, it is the middle part, and the sides are the extremes
disjunct.
sin B 0 = cos A B . cos A 0;
taking the complement of the hypothenuse,
cos B 0 = cos A B cos A 0;
SPHERICAL TRIGONOMETRY. 31
log cos B 0 = log cos A B -f- log cos A 0 — 10
= 9*8691301 + 9-7872317 - 10
= 9-6563613, which is the cosine 62° 31',
which is less than 90°, A 0 and B 0 being alike.
When a side and its opposite angle are given.
Given the side A 0 = 55°, and its opposite angle B =
65° 46' 5", to find the remaining parts of the triangle.
To find the other angle 0.
Here B is the middle part, being separated from A C by
A B, and from the angle 0 by B C;
.*. A 0 and 0 are the extremes disjunct.
sin B = cos A 0 cos C;
taking the complements of B and C,
cos B = cos A 0 sin C;
log sin 0 == log cos B — log cos A 0 + 10
= 9-6132407 + comp. log. 0-2414087 + 10
= 9-8546494 = log sin 45° 41' 21"
/. C = 45° 41' 21".
The angle C is ambiguous; as it cannot be determined by
the data alone whether A B, C, and B 0 are greater or less
than 90°.
To find the side A B.
Here AB is the middle part, AC and B the extremes
conjunct.
sin A B = tan A 0 tan B;
taking the complement of B,
sin A B = tan A 0 cot 3 ;
log sin A B = log tan A 0 + log cot B — 10
= 10-1547732 + 2-6532976 -10
= 9*8080708, which is the sin 40°;
A B = 40°.
32
SPHERICAL TRIGONOMETRY.
The side AB is also ambiguous for the same reason as
above.
To find the hypothenuse B 0.
The side A 0 is the middle part, and B 0 and B are the
extremes disjunct.
sin A 0 = cos B 0 . cos B;
taking the complements of hypothenuse and angle, B,
sin A 0 = sin B 0 sin B;
log sin A C = log sin B 0 + log sin B — 10
log sin B C = log sin A 0 — log sin B + 10
= 9*913364:5 + 0*04:00568 + 10
= 9-9534:213;
B 0 = 63° 56' 7".
When a side and its adjacent angle are given.
Given the side A 0 = 54/ 4:6', and its adjacent angle
47° 56', to find the remaining parts.
To find the side A B.
Here the circular parts all lie together, hence A 0 is the
middle part, and A B and 0 the extremes conjunct.
sin A 0 = tan A B tan 0;
taking the complement of 0.
sin A 0 == tan A B cot 0 ;
log A 0 = log tan A B + log cot 0 — 10
log tan AB = log sin A 0 — log cot 0 + 10
= 9-9121207 - 9-9554535 + 10
= 9*9566672 which is the tangent of 42° 8' 46";
" /. AB = 420 8'46/',
which is acute, like its opposite angle.
To find the angle B.
Here B is separated from the two given quantities; calling
it the middle part, then A 0 and 0 are the extremes disjunct.
SPHERICAL TRIGONOMETRY.
33
sin B = cos A 0 cos 0;
taking the complements of B and 0,
cos B = cos A C sin 0;
log cos B = log cos A 0 + log sin 0 — 10
= 9-7611063 + 9-8706179 - 10
= 9-631724:2, which is cos 64:° 38' 31";
•. B — 64:° 38' 31".
To find the hypothenuse B 0.
Here the circular parts all lie together, and 0 being in the
middle, is the middle part, and B 0 and A 0 the extremes
disjunct.
sin 0 = tan B 0 tan AO;
taking the complements of the hypothenuse and of angle 0,
cos 0 = tan A 0 cot B 0;
log cos 0 = log tan A 0 + log cot B 0 — 10;
log cot B 0 = log cos 0 — log tan A 0 + 10
= 9-8260715 - 10-1510145 + 10
= 9*6750570, which is the cotangent o£
64:° 4:0' 34/';
.-.BO = 64:° 40' 34//.
QUADKANTAL TRIANGLES.
39. Quadrantal triangles can be solved by the same rules as
right-angled triangles for using the polar triangle; we see that
since
one side is a quadrant, and that in the polar triangle
A/= 180 — a;
.-. A'=180° - 90° = 90.
In the polar -triangle, since A' have by the equa¬
tions, page 10,
cos a' = cos b1 cos d
sin b1 = sin a' sin B'
tan b1 = tan a' cos C
sin c' = sin a1 sin C;
tan cf = tan a' cos B'
o3
34
SPHERICAL TRIGONOMETRY.
tan V = sin c' tan B'
tan c' = sin V tan 0'
cos B' = sin 0' cos V
cos 0' = sin B' cos d
cos a' = cot B' cot c'
From these by substituting these values
a! = 180° - A; V = 180° - B; c' = 180° - 0;
A' = 180° - a; B' = 180° ~ 6; 0' = 180° - c;
we get these results,
cos A = — cos B cos C
sin B = sin A sin b
sin 0 = sin A sin b
tan B = — tan A cos c
tan 0 = — tan A cos I
tan B = tan b sin c tan 0 = sin B tan c
cos b — sin c cos B cos c = sin b cos 0
cos A = cot b cot c
Or without using the polar triangle,
.
cos a — cos b cos c
cos A = :—— make a = quadrant,
sm o sin c
^
'
then cos a = 0, and we have
» cos b cos c
cos A
cot b cot c:
:—-—:
sin 6 sm c
'
i}
cos b — cos a cos c cos b
cos B =
sm a sm c sm a
cos c — cos a cos b
cos c
cos 0 =
sin a sin b sin b
From these equations, and the equation S?n ^ = s^n a We
sin B
sin V
can deduce all the cases of quadrantal triangles.
Given A B = c = 32° 57' 6" and A 0 = b = 66° 32' to
find B and A,
'
cos A = — cot b cot c
log cos A = log cot b + log cot c — 10
= 10-1882850 + 9-6376106 - 10
= 9*8258956, which is the cosine of 47° 57' 16"
but since cos A is negative, A must be greater than 90°.
SPHERICAL TRIGONOMETRY.
35
OBLIQUE-ANGLED TRIANGLES.
40. Case 1. Given the three sides, viz.
AB
= 79° 17' M" \
BO = 110° (to find the rest.
AC
=r 58° j
To find the angle A.
sin
- 6 ) sil1 ("-").
V sin b sin c
'
hence we have the following rule :
From half the sum of the three sides subtract each of the
two sides which contain the required angle.
Add the log sines of these two remainders, and the arith¬
metical complement logs of the sines of the sides which
contain the angle.
Half the sum of these four logarithms will give the log
sine of half the required angle. Thus:
79° 17' 14"
110
58
By page 18, sin J A =
A/
2)247 17 14
123 38 37 = J sum of the three sides.
79 17 14
44 21 23 first remainder log sin = 9,8445513
123 38 37
58
65 38 37 second remainder
log sin = 9,9595173
comp log sin 58° 0,0715795
comp log sin 79° 17' 14/' 0,0076359
2)19,8832840
log sin = 60° 57' 28" = 9,9416420
2
121 54 56 equals the required angle A.
36
SPHERICAL TRIGONOMETRY.
By a similar operation the angles B and 0 may be found;
but when one angle is known, the other two are easily deter¬
mined by Art. 13, page 6.
Case 2. Given the angle A = 32° 20' 30", the side
5 = 72° 10' 20", and the side a = 78° 59' 10", to find B, 0
and c.
tt ,
. sin 6
Here
by page•
6, sm Br, = sin A;
sm a
log sin B = log sin A + log sin b — log sin a
log sin A = 9*7283269
log sin b = 9*9786283
19*7069552
log sin a = 9-9919261
log sin B == 9*7150291
B = 31° 15' 15"
By page 9, equation (16)
cot J2
O = tan J v(A n
+ B)J CQS % (a +
cos | (a — b)
log cot J 0 =
log tan § (A + B) + log cos J (a + b) — log cos J (a — b)
log tan J (A + B) = log tan 31° W 52" = 9-7217470
log cos J (a +5) •-= log cos .75° 34/ 45" = 9*3962727
log cos -J- (a — 6) = log cos 3° 24/ 25" =
log cot J 0 =
... jo = 82° 30' 39";
or O = 165° 1' 18".
19-1180197
9*9992318
9*1187879
We might find c from the equation
sin C
sin c = sm a . -—r,
sm A
but we can find it directly from Napier's Analogies.
SPHERICAL TRIGONOMETRY. 37
By page 9, equation (14:), we have
tan
.
c =
tan i(a + b).
;
log tan 50 =
log tan J (a + 6) + log cos i (A + B) — log cos |(A — B)
log tan J (a + 5) = log tan 75° 34/ 45" = 10-5898236
log cos | (A + B) = log cos 3104:7' 52" = 9-9293745
20-5191981
log cos i (A — B) = log cos 1° 5' 15" = 9-9999218
log tan 50 = 10-5192763
i c = 73° 10' 10" '
c = 146° 20' 20".
Case 3. Given 0 = 30° 45' 28"; a = 84° 14' 29"; b =
44° 13' 45", the two sides and the included angle, to find
A, B, c.
By Napier's Analogies, page 9, equations (16) and (17),
tan J2 (A
= cot}3 C . 003 f ~ ^
K + B)'
cos i (a + 6)
and tan i
= cot ^ 0 . 8laf{a~^
2 (A
v — B)
^
sin 5 (o + 6)
| 0 = 18° 22'44" log cot = 10-4785395
i (o - 6) = 20° 0° 22" log cos = 9-9729690
20-4515085
£ (a + 6) = 64° 14' 7" log cos = 9-6381663
log tan J (A + B) = 10-8133422
J (A + B) = 81° 15' 44" • 41.
3& SPHERICAL TRIGONOMETRY.
A — B determined,
J 0 = 18° 22' U"
J (a—b) = 20 0 22
log cot = 104785395
log sin = 9*534:1789
J (a + 6) = 64: 14:
20-0127184:
log sin = 9-9545255
7
10-0581929
J (A - B) = 48° 49' S8".
A and B determined. c determined,
i (A+B)=810 15' 44:,,-41 log sin 36° 45' 28/,= 9-7770158
!(A-B)=48 49 38
logsin44 13 45 = 9-8435629
A = 130° 5' 22"-41 19-6205787
B = 32 26 6-41 log sin 32 26 6 =*= 9-7294422
log sin c = 9-8911365
c = 51° 6' 121',
c may be found directly, without finding A and B, by the
following method:—
cos c—cos a . cos ft ^
Since cos 0 = : :—7 , page 5t
sin a . sin o
.\ cos c = cos a. cos b + sin a. sin 5. cos 0;
but cos 0 = 1 — ver. sin C,
cos c = cosa. cos b + sin a . sin 5—sin a. sin b. ver, sin
= cos (a—b) —sin a . sin b . ver. sin C;
... 1—cos c, or 2 sin2-|'=ver.sin(a-—5)+sin a. sin b. ver.sin
= ver-
. ,
/
sin a . sin 6 . ver. sin 0 \
\ 1 + ver. sin (a—b) }
sin a . sin b . ver. sin 0
ver. sin (a—6)
8,n
SPHERICAL TRIGONOMETRY* 39
which in logarithms is 2 log tan 0 =
log sin a+log sin 5+log ver. sin 0—log ver. sin (a—b)...[a]
then 2 sin2-^- =ver. sin (a—b) . sec2 0, and
log 2.+21ogsin-^-=log ver. sin (a—6)+2 log sec 0—10...[6]
c computed independently of A and B.
Finding the auxiliary angle 0 by the foam [a],
a = 84;° 14! 29" sin = 9-9978028
5 = 36 13 45
sin = 9-8435629
0= 36 45 28 ver. sin = 9-2984762
o — 6 = 40
0 44
.-. 2 log tan 0
and log tan 0
29-1398419
ver sin = 9*3693878
= 19-7704541
= 9-8852270
Case 4. Given c = 50° 6' 20"; A = 129° 58' 30";
B = 340 29' 30"; to find a, b, 0.
By equations (14) and (15), page 9,
tan ^ (o -f^ c . 008 i
5}
T 6) ^ tan 2
'
cos£(A + B)
tan £ (a - b) = tan £2 c . S|n f (A ~ B)
^
'
sin £ (A + B)
£ (A + B) = 82° 14'
\
i(A - B) = 47° 44' 30" (
| c = 25° 3' 10")
log tan i (a + b) =
log tan J c +log cos|(A — B) — logcos|(A + B)
log tan i c = log tan 25° 3' 10" = 9-6697162
log cos J (A—B) = log cos 47° 44' 30" = 9*8276758
log cos ^ (A + B) = log cos 82° 14'
log tan J (a -f b)
19-4973920
= 9-1307812
= 10-3666108
40
SPHERICAL TRIGONOMETRY*
••• i (a + 6) = 66° 4:4:' 10
log tan J (a—6) =
log tan J c + log sin J (A — B) — log sin £ (A + B)
log tan i c = log tan 25° S' lO"
= 9*6697162
log sin|(A—B) = log sin 47° M' 80' = 9*8693023
log sin £ (A + B) = log sin 82° 14'
19-5390185
= 9*9959977
log tan £ (a—5) = 9*5430208
i (a-b) = 19° 14' 50'7
J (a + b) + J (a — b) = a
£ (a + b) — £ (a — b) == b
66° 44' 10"
19° 14' 50"
85° 59' 0"
47° 29' 20"
a = 85° 59' and b = 47° 29' 20".
To find 0.
sin A
sin a
. «
.
or, sin U = sin c
sin G
sin c
sin A
. ~—
sin a
log sin 0 = log sin c + log sin A — log sin a
log sinA=log sin 129° 58' 30" = 9*8844129
log sin c = log sin 50° 6'20" = 9*8849241
log sin a = log sin 85° 59'
Jog sin 0
19-7693370
= 9*9989319
= 9*7704051
/. 0 = 36° 6' 50".
SPHERICAL TRIGONOMETRY. 41
Oase 5. Given the angles A = 70° 39'; B = 48° 36';
a = 89° 16' 53", to find the rest.
By Art. 33, page 21, we have sin 6 = 8111 a
log sin b == log sin a + log sin B — log sin A
log sin a = log sin 89° 16' 53" = 9-9999658
log sin B = log sin 48° 36'
= 9*8751256
19-8750914:
= 9*974:7475
log sin A = log sin 70° 39'
log sin 6 = 9*9003439
6 = 52° 39' r
sin 5 =sin (180 — b)
sin 127° 20' 56"; but since A > B,
a must be greater than 6, hence b cannot be 127° 20' 56'/.
t
To find c.
By Napier's Analogies
tan
003
c = tan £(«
f^
^
2 v + &).
; cos i
(A — B)
log tan J c =
log tan i (a + b) + log cos | (A + B)—log cos i (A — B)
log tan | (a + b) = log tan 70° 57' 59* = 10*4622011
log cos | (A + B) = log cos 59° 37' 30" = 9*7038563
201660574:
log cos £ (A — B) = log cos 11° 1' 30" = 9-9919097
log tan | c
= 10-1741477
£ c = 56° ir 29",
or c = 112° 22' 58/,.
By equation (16), page 9, we have
cot 10 = tani(A+B)
^
cos ^ (a — ft)
**
SPHERICAL TRIGONOMETRY.
log COt ^ 0 log =
log tan J (A + B) + log cos J (a + b) — log cos J {a—b}
log tan £ (A -f B) — log tan 59°- 37' 30" = 10-2320208
log cos i (a + b) log cos 70° 57' 59" = 9-5133811
19-74:54019
log cos £ (a — b) = log cos 18° 18' 54" = 9-9774233
log cot J 0 = 9-7679786^ 0 = 59° 37' 30",
or 0 = 119° 15'.
Case 6. Given the three angles,
angle A = 120° 54' 56" \
angle B = 50 >res^
angle 0 = 62 34 6 ) •
By page 22, cos * «=
/cos (S-B) cosJS-C>;
V sm B sm 0
hence the following rule.
To find the side B CL
From half the sum of the three angles take each of the
angles next the required side.
Add the log cosines of these two remainders, and the
cpmp. log of the sines of each of the adjoining angles.
Half the sum of these four logarithms will give the cosine
of half the required side; thus
SPHERICAL TRIGONOMETRY.
43
121° 54:' 56"
50
62 34 6
2)234 29 2
117 14 31
62 34 6
54 40 25 first rem. cos 9,7621032
67 14 31 second rem. cos 9,5875321
comp. log sin 50° • 0,1157460
comp. log sin 62° 34' 6" 0,0518018
2)19,5171831
cosine 55° 9,7585915
2
110° = the required side B 0.
By the same method the other sides may be found; but
one side being known (with the angles) the rest are most
readily found by Art. 14, page 6.
44
SPHERICAL TRIGONOMETBY^
CHAPTER III.
41. The surface of the sphere included between the arcs
D M, D N is proportional to the angle N D M or the arc
M N. See fig. page 1.
If the circumference be divided into equal parts as M N,
and great circles be drawn from D through the points M, N,
the portions of the surface, such as N D M, are all similar and
equal, hence if F M contains N M, n times, or if F M = n
times N N, the surface F D M will be n times N D M.
When D M coincides with D G, the angle FD G- or its mea¬
sure F M Gr =180; hence if S = whole surface of the sphere,
and if A be the angle N D M which is measured by the are
S A
N M, the surface N D M = -j •
but S = area of 4 great
circles of the-sphere. Hall's Diff. Cal., page 370.
S = 4 tt r2 = 4 tt when radius is unity,
or y = tt = 180.
4
The measure of the surface of a spherical
trangle is the difference between the sum
of its three angles and two right angles.
Let the triangle be A B C, a, 6," c, repre¬
senting the magnitudes of the angles at
A,B,C;let P=surfaceBOiB,Q±=mCwm,
R = AO n A; produce the arcs 0 m, 0 n,
till they meet at e (which will be on the
hemisphere opposite to that represented by
A B mnA)% then each of the angles at 0 and
e
e equals the angle of the planes in which the arcs C m e, C n ey
lie; therefore the angles at 0 and e are equal.
SPHERICAL TRIGONOMETRY.
45
Again, the semicircles A 0 m, 0 m e; B C nt 0 n e are equa
or, AG-f-OTTi = Cm + me,and/. A 0= me,and BO =nc
aud the triangle men — the triangle A B 0; let x = its area
then, by last article,
x+
2
180
*+Q=! • ^andz+P + Q + R^
a: + R=— •
consequently, by addition,
2 * + <a: + P + Q+ R) or 2a: + | = | . ^d±±cj
*' X = 4 (180) ^
j=a+&+c — 180°;
or r* (a + 6 + c — 180°).
Hence the area of a spherical triangle is equal to the ex¬
cess of the sum of its three angles above two right angles,
which is called the spherical excess.
The late Professor Woodhouse, in his able work on Tri¬
gonometry, observes that—" This expression for the value
of the area was merely a speculative truth, and continued
barren for more than 150 years, till 1787, when General Eoy
employed it in correcting the spherical angles of observation
made in the great Trigonometrical Survey."
In a biographical sketch of the life of Isaac Dalby, late
Professor of Mathematics at the Eoyal Military College,
Sandhurst, in Leybourne's Mathematical Repository, it is
stated that he had sent some years previously to his death
an account of the principal events of his life after reaching
maturity. The following is a quotation from himself given
in the above-named excellent periodical:—
"General Koy^ account of this measurement is in the
Philosophical Transactions ; but it is not altogether what it
ought to have been. His description of the apparatus, de¬
tail of occurrences, &c., are all well enough; but he should
not have meddled with the mathematical part, for Ms know¬
ledge did not extend beyond Plane Trigonometry. I drew up
^
c —
)
46
SPHERICAL TRIGONOMETRY.
the computations in that form which I thought the most pro¬
per for publication, but he was continually making alterations.
He didnot even understand the rule I made use offor finding
the excess of the sum of the three angles of a spherical triangle
above 180° (which since that timehas been quoted as General
Roy's theorem), and would not insert it until he had consulted,
the Hon. Henry Cavendish. For conducting the business in
the field, however, few persons could have been better qua¬
lified than the General. . .
" I believe he was the best topographer in England, and
knew the situation of every barrow, cairn, and hillock in
Great Britain. He had something of an observatory in the
upper part of his dwelling, and could regulate a clock or
watch by means of transits. In fact, he was ready enough at
calculations which depended merely on the use of the tables.
But the rules which he published for measuring the heights
of the barometer all came from Mr. RamsdenV
A note is given to this extract in the Repository, which is
as follows:—" It is not until very recently that Mr. Dalby
has had justice done him with regard to this ingenious rule.
At page 138 of the new edition of Yol. III. of Hutton's
Course of Mathematics, published by Dr. Gregory in 1827,
we find this note :—' This is commonly called General Roy's
rule, and given by him in the Philosophical Transactions for
1790, p. 171; it is, however, due to the late Mr. Isaac Dalby,
who was then General Roy's assistant in the Trigonometrical
Survey, and for several years the entire conductor of the
mathematical department.'"
FURTHER DEVELOPMENTS CONCERNING THE SPHERICAL
EXCESS.
42. Let the radius of the sphere be unity, tt the semicircumference of a great circle; a, &, c, the three sides of a
spherical triangle; A, B, 0, the arcs of a great circle that
measure the opposite angles.
Let the spherical excess A + B + C— 7r=S.
The area of the spherical triangle is equal to the arc.S
multiplied by the radius, and is therefore represented by S.
Now, by Napier's Analogies,
SPHERICAL TRIGONOMETRY.
47
tan I (A
+ B) = cot £ 0 . —^
- v 7
^
cos f (a + 6)
an
/A
T»\
"I" ^an "i" ^
tan f 1 (A
+ B)
+ J 0 })
= ^1 _-"f"
tan ^ (A+B) tan ^ c
cot i 0 .
C0S
r I""?? + tan 4 0
cosJ(a + 6)
1—cot!2 0 . C03|(a-°j . tan i 0
cos 5 (a+6)
(a ~ H + tan 10
cos J (a + 6)
^ cosJ-(a — b)
cos -g" (& -j- 5)
cot i 0 .
*
_cot4- 0 . cos J (a — 6) + tan ^ C . cos J (a + 6)
~~ cos ^ (a + b) — cos J (a — 6)
1+cosC 1—cosO
,,
—=—cos i{a—b)
^ cos i (a + o)N
2 v
yH .
sm 0
sm 0
a\ » /
cos % (a + 6) — cos -J- (a — 6)
(by expanding the cosines and reducing)
1
cos J a cos | b + sin % a sin J b cos 0
sin 0 — sin ^ a sin J b
— cat ^ a cot | b — cos 0
sin 0
but tan f (A + B + 0) = tan i (S + 180) = — cot J S ;
^ „
cot i a cot i b + cos 0
cot J^
S=
oin ( :
4:8
SPHERICAL TRiaONOMETRY.
This equation, which is very simple, enables us to find the
area of a spherical triangle when the two sides and the in¬
cluded angle are given.
To find the area of a spherical triangle in terms of the
three sides,
, , ^
cot i a cot -J- b + cos 0
cot I S = = .
'
sin 0
^ cos c — cos a cos b
COS 0 = ; ;—
sin a sin 6
1-f cos a A ^ ^ l-^cos b
and cot JJ o = T
, cot J b
oir*
* 8111 b
sin a" '
1+cosa + cos6 + cosc
COS 0 + cot -k a cot bb~
: : 7
*
*
sin a sin 6
_ . a4-6+c . a2-sin
sm
^ cos c—cos (a+6)
1 + C0SC =
: r—jr sin a sin b sin a sin b
hb c
2sm(a + c—b) . b+c—a
/
<7 c
^-77
. _ cos (a—6)—cos
2 sin2
1 —COS C =
^^^ = ; : 7
sin a sin o sin a sm b
Multiplying these two equations and. extracting the root,
/To -4-6 + c . a-\-b—c . a-fc— b . 64-c—a
sm — sm
A / sm —— sm
sin C = 2
sin a sin b
By substituting these values we have
cot -J- S = 2
1 + cos a + cos b + cos c
4-&4-C
. a+6—c . a+c—6 . 6+c
/. a 1
sin
sm—-—sina / sm— 2
2 2 2
a
This solves the problem, but it can be put into a simpler
form, and one that is adapted to logarithmic computation.
SPHERICAL TRIGONOMETRY. 49
Kesuming the formula
fiq
COt4b=
cot! a cot J 6 + cos 0
siiTO
'
l + cot2|S=:T 21
sin JS •
cot2-Ja cot21-6 + 2 cot J a cot £ b cos O+l
£in2 0
'
by multiplying botli sides of the value of cos 0 by 2 cot J a
cot J 6, we have
COS C COS
2 cot. I1 a coti ■£i . b7
cos 0 =
.d COS 1} ,
2 sin2 -I a sin2 J 6
putting in
the numerator for cos c, cos a, cos b, their values
1-2 sin2 J c, 1 — 2 sin2 Ja, 1 — 2 sin2 J 6, we shall have
2 cotiacot^
cos C =
2
3
+
sm2^asin2|6
Also, cot2 -J-a cot2 J 2
"■ 1~S22if^5
sin fa
sin J6
1 — sin2 \ a — sin2 ^ b , ^
sin2 i a sin2 b b
Substituting all these values, we have
1—
1 — sin2 \ c
or
sin J S
sin? ^ a sin2 J b sin2 0'
2
. i r* sin -J- a sin i b sin 0
sin -I S = ^
cos c
Substituting for sin 0 its value, we have
D
2_
50
SPRTTRICAL TRIGONOMETRY.
ein ^ S =
. a + ^i-c . a + 6 —c. a + c--b . b + c—a>
2 cos
a cos J 6 cos J c
which expression is adapted to logarithmic computation.
If we multiply this equation by cot J S we have
COS ^ S :
1 + cos a + cos b + cos c
4 cos ^ a cos J b cos J c
cos2 ^ a + cos2 J b + cos2 J c—1 ^
2 cos ^ a cos 6 cos J c
'
From this we have,
1 — cos i S A , 0
. , q
or tan J S =
sin J S 4
1—cos2*! a — cos2 J &—cos2-I c + 2 cos J a cos J 5 cos \ c
/ f . a+&+c . a+6—c . a+c—6 . 54-c—a")
V t8m~2
'Bin
2
'8111
2
'sm 2 /
The numerator of this expression can be put under the
form (1—cos2 ^ a) (1 — cos2 J b) — (cos J a cos ^ b — cos J c)2
which may be decomposed into the factors
sin J a sin ^ b + cos J a cos J b — cos J c and
sin £ a sin \ b — cos % a cos J b + cos J c.
These reduce ultimately, the first to cos (-J a —^ b)—cos J c=
. a+c — 5sm
. 5 + c—a
20 sm—
#
4 4
'
the second to cos J c — cos (J a + ^ &) =
^2 sin
. a-J-fr-j-c
• —-& — ;
c
, - sm
t
SPHERICAL TRIGONOMETRY.
. .
4: sm
51
tan i S =
— C . Cl-J-C—— & » S-f-C"-®
am
sin
sm
/ ( . a+S+c . a + 6—c . a+c—b . 54-c—a)
Sln
sm
8m
V rn
2
2
2
/
. tan £ S =
a4-6+c
a+b—c
a+c—b
&+c—a
tan
tan
tan
. tan
•
4 4 4 4
This elegant formula is due to Simon Lhuillier.
Legendre's Geometry, page 319.
See
GIKARD'S THEOEEM.
43. By page 45,
ISO0
a 4- b + c — 180° = r x, or reducing a, 6, c to seconds,
TT T*
180° x 60 x 60
the excess m seconds =
5
x.
TT H
Now, on the earth's surface, the length of 1° taking a
mean measurement = (60859*1) x 6 feet, and an arc =
' 360
radius = -— ;
A TT
360
.*. (60859*1) x 6 x —— = radius of the earth in feet;
A TT
.
,
2 ir
60 x 60
the excess m seconds = x . r-— • 2
360 6 x (60859-1)2
2s- x 10
.x.
36 x (60859-1)2n'
log . excess =
log x —12 (log 6 -4- log 60859-l)—log 2 tt x 10^
= log * - (11-1249536 - 1-7981799)
= log x - 9-3267737.
d2
52
SPHERICAL TRIGONOMETRY.
Hence the following rule:—
From the logarithm of the area of the triangle, considered
as a plane one, in feet, subtract the constant logarithm
9,3267737, and the remainder is the logarithm of the excess
above 180°, in seconds nearly.
Observed angles.
Ex. Hanger Hill Tower .
(а) 42° 2' 32"
Hampton Poor-house,
(б) 67 55 39
King's Arbour
(c) 70 1 48
Distance from (a) to (b) = 3,84:61*12 feet, .
from (a) to (c) = 24704:-7.
Taking the distance from (a) to (c) for the base of the
triangle, the //perpendicular on the base will be 38461 *12 x
sin 4:2° 2' 32 , and therefore the area of the triangle
base
perpendicular
2
= 24:704:-7 x 19230*56 x sin 42° 2' 32",
log area = log 24704: 7 + log 19230*56
+ log sin 42° 2' 32" - 10
= 4-3927761 + 4-2839906 + 9-8258661
= 8*5026328 = logarithm of the area in feet ;
hence, 8*5026328 — 9*3267737 = — 1*1758591 ,*
the corresponding natural number is *14992, the spherical
excess in seconds.
legendre's solution of spherical triangles whose sides
ARE VERY SMALL COMPARED WITH THE RADIUS OF THE
SPHERE.
44. When the sides a, b, c, are very small with respect to the
radius of the sphere, the proposed triangle is very little dif¬
ferent from a rectilinear triangle, and, considering it as such,
we can have a first solution approximately, but we neglect in
this manner the excess of the sum of the three angles above
180°. To have a solution more approximate, we must take
SPHERICAL TRIGONOMETRY. 53
into account this excess, and this we can do very easily by
means of a general principle which we proceed to demon¬
strate.
Let r be the radius of the sphere upon which the triangle
is situated, and if we imagine a similar triangle upon the
sphere whose radius is unity, the sides of this triangle will
be -, —, —; and we shall have cos A =
r r r
a be
cos — — cos - cos r r
r
. b . c
'
sm - sin —
r
r
but since r is very great with respect to a, b, c, we shall have
approximately,
a a2 <n'4
COS r
- = 12r
- o-~2 + 2.3. ir4'
b
, .
&<
cos — = 1 — 2
r 2r
2.3.4r4
c c2 c4
COS - = 1 - 2 ;
r 2r
2.3 .Ir4*
•
8111
- -t- '_^L f
r
r" 2.3^
. c
c
sin — =
r
r
2
'
Substituting these values in the above equation
+ c2 _ a2
a4 ^^4 __ c4
&2 c2
2r2
24: r4 4:r4
cos A=
——
6c
62
c2 \
b c /A
2
2
r \
6r
6 r2 /
2)2 i 2
Multiplying
lying numerator
numeral and denominator by 1 -j
and
6 r2
reducing
54:
SPHERICAL TRIGONOMETRY.
COS A =
b2 + c2 — a* a* + bA + c* — 2a2b2 — 2a2c2 — 2&2c2
2 6c
24: 6 c r2
Let A' be the angle opposite to the side a in the recti¬
linear triangle of which the sides are equal in length to the
arcs a, 6, c, we shall have
cos A' — ^ + <f —
Abc
a2
, and
4: 62 c2 sin2 A' = 2 a2 62 + 2 a2 c2 +2 62 c2 — a4 — 64 — c4;
6c
therefore cos A = cos A' — tt-o . sin 2A!:
o r*
let A = A7 + x, we shall have rejecting the square of
cos A — cos A' — x sin A',
from whence we have x =
6r2
s n
i A',
6
c
and since # is of the second order with respect to - and it
r
r
follows that the result is exact to quantities of the fourth
order, we shall then have *
A = A' + ^2 sin A';
br
but ^ b c sin A' = the area of the rectilinear triangle, of
which the three sides are a, b, c, do not differ sensibly from
those of the proposed spherical triangle. Then, if either
area be called a, we shall have
A=A' + ^2) orA' = A-^;
Similarly, B'= B — ^; 0' = 0 — ^;
hence there results,
SPHERICAL TRIGONOMETRY.
55
A' + B' + C or 180 = A + B + 0 — -Jj,
we "can then consider
as the excess of the three angles of
r
the spherical triangle above two right angles.
Hence we have the following rule :—
A spherical triangle being proposed, of which the sides are
very small with respect to the radius of the earth, if from
each of its angles one-third of the excess of the sum of its
three angles above two right angles be subtracted, the angles
so diminished may be taken for the angles of a rectilinear
triangle, the sides of which are equal in length to those of
the proposed spherical triangle; or, in other terms :—The
spherical triangle, whose sides are nearly rectilinear, of which
the angles are A, B, 0, and the opposite sides a, b, c, answer
always to a rectilinear triangle whose sides are of the sarm
length, a, b, c, and of which the opposite angles are A —^ e ,
B — ^ t; 0 — -jc; e being the excess of the sum of the angles
of the spherical triangle proposed above two right angles.
The excess t, or
wdiich is proportional to the area of the
triangle, can always be calculated d 'priori, by the given parts
of the spherical triangle considered as rectilinear. If the
two sides, b, c, and the included angle A, are given, we shall
have the area a = b c sin A; if we have given the side a, and
the two adjacent angles B, 0, we shall have the area
sin B sin 0
8in (B + C) *
45. Given the three edges of a
parallelepiped, and the angles be¬
tween them, to find the solidity.
Let the edges S A = /, 8 6 = #,
SC = h, and the contained angles
ASB = a, A SC = g, BSC =
y, if from the point C we let fall
C 0 perpendicularly on the plane
ASB then in right-angled triangle
OSO; CO = OS sin CSO =
h sin 0 S 0, besides the surface of the
56
SPHERICAL TRIGONOMETRY.
parallelogram A S B P =fg sin a. Therefore, if we call S
the solidity of the parallelopiped S T, we shall have S =
/. g. Ti. sin a sin 0 S 0. We now proceed to find sin OSO.
From the point S as a centre and radius unity, describe a
spherical surface meeting the right lines S A, SB, SO, SO, in
the points D, E, F, G, we shall have a triangle D E F, in
which the arc F G is perpendicular to E D, since the plane
0 S 0 is perpendicular to A S B. Now, in the triangle D EF,
where the three sides, D E = a, D F = €, E F = y, we have
^
COS € — COS a COS y
cos E =
: : and
sm a sm y
sin E =
— COS2 a — cos2 6 — cos2y + 2 cos a cos € cosy
sin a sin y
Then, in the right-angled triangle EFGr, sin G-F or sin OSO =
sin E sin E F == sin y sin E,
S = /. g. A. sin a sin y sin E =
f.g.Ji. */1 — cos2 a — cos2 € — cos2 y + 2 cos a cos € cos y.
The expression under the radical is composed of the two
factors, sin « sin y + cos € — cos a cos y, and sin a sin y —
cos € + cos a .cos y, the first
cos € — cos (a + y) r=
2 sin a-f-€,4"Y .
2
'Sin
a
-{- y — €
;
2
the second == cos (« — y) — cos € =
2 sin
«'+€ — y sin
. €+ y—a
^ •
2 '
therefore, S =
. a+£+y
. a+€—y Sin
. «+y—€ 3111
. €+y —«
1
Sln
g
~2—
2 —'
zfgh's/1sin
46. The same things being given as in the above to find
the diagonal.
SPHERICAL TRIGONOMETRY. 57
Let the diagonal of the base S P = z, and the required
diagonal S T = u, the triangle A S P, iii which cos S A P =
— cos «, we have z2 =f2 + g2 + 2fg cos a, in like manner
in2 the 2triangle
T S P, in which cos T P S = — cos 0 S P;
u = z + h2 + 2 h z cos 0 S P.
We must now find cos 0 SP, or of the arc F H.
Now, in the spherical triangle EFH, we have cos F H =
cos E F cos E H + sin E F sin E H cos E, substituting the
• _ _
^ COS € COS a cosy- ,
values F F = y and cos E = ; ; it becomes
'
sm a sm y
^TT
-»-• EM
tt i sinEH
.
^
€03 F
H = COS y COS E
; (cos
b — COS a COS y). =
'
sm a
sin E H cos € ^ sin (a — E H) cos y _
sin a sin a
sin E H cos € + sin D H cos y
sin a
Therefore 2 hz cos F H, or 2 h z cos 0 S P =
h, cos -pb z sin E H
+, 20 ti,
cos y .^ sin;D H
:
sin a sm a
but in the triangle B S P we have
0
2
BF= 8 P.^P, .»dBS=S F »» »18,
sin S B P sm S B P
„which
i • •i gives
' n z sin E=
H /r and,
, z sin D=
H
q:
:
sm a
* sin a
/. 2 h z cos 0 S P = 2 fh cos € + 2 g h cos y<
w2
Hence the square of the required diagonal •
^ g2 ^ ^2 _J. 2 fg COS a + 2/A cos € + 2 g h cos y.
47. To determine a line on the surface of a sphere on
which the vertices of all triangles of the same base and sur¬
face are situated.
d3
58
SPHERICAL TRIGONOMETRY.
Let A B C.bea spherical triangle, (one of those on the com¬
mon base A B, = c) ; and the given surface A + B + C —tt
= S. Let I P K be an indefinite perpendicular on the middle
of A B, having taken IP = a quadrant, P will be the pole
of the arc A B,and the arc CD, drawn through the points
PC, will be perpendicular to AB. Let I D ==p.C D = q,
the right-angled triangles, A C D, B C D, in which AC =b,
BC = a, AD=p + -|c, BD=p — Jc, will give cos a
=: cos q, cos (p — c), cos b = cos q cos (p + jtc). But it
was found, page 48, that
1 4- cos a 4b 4- cos.
c
cot. -i"■, aS = ■——
T cos
7 . '
sm a sm 6 sin C
Substituting in this formula the values
cos a -j- cos 6 — 2 cos q cos p cos c,
1 — cos 0=2 cos2 i c, sin b sin 0 =
sin c sin B = 2 sin J c cos ^ c sin B; a
there results,
cot, J, cS = cos
—r*i c 4T cos rp. cos q
sin a sin c sin B
Again, from the right-angled triangle B C ]}, sin a sin B =
sin q;
cos ^ c 4- cos p . cos q
cot A S
sm c sin q
or, cos p cos q = cot J S sin c sin q — cos J c. This is the
relation between p and q which will determine the locus of
all the points C.
Produce I P to K, let PK = x. Join KC, andlet KC =y;
in the triangle P K C where we have P C = ^ tt — q, the
angle KPC = tt — p, the side KC will be found by the
formula
cos K C = cos KPC sin PK sin P C -|- cos P K cos P C, or
cos y = sin q cos x — sin x cos p cos q.
Substituting this instead of cos p cos q the value
coa | S sin i c sin q — cos | c, there results,
cos y = sin x cos c -f- sin q (cos x—sin x cot iV S sin ^ e),
SPHERICAL TRIGONOMETKT.
59
in which, if we take cos x — sin a; cot J S sin ^ c = 0, or
cot ^ S sin ^ c cot x, there will result cos y = sin x cos J c,
and thus a constant value of y is determined.
Therefore, if after having drawn the arc IP perpendicular
to the middle of the base AB, and beyond the pole the part
P K such that cot P K = cot \ S sin ^ c, all the vertices of the
triangles on the same base c, and of the same surface S, will be
situated on the small circle described from K as a pole at the
distance K C, such that cos K C = sin P K cos \ c. This is
Lexell's theorem.
48. Given the three sides, B C = a, A 0 = b, A B == c, to
find the position of the point I, the pole of the circle cir¬
cumscribing the triangle ABO.
Let the angle ACI = #,and the arc AI = CI = BI =0;
fn the triangles 0 A I, C B I, we shall have by the equation—
cos 0 — cos b cos 0
cos x = :—;—:—
sin o sin 0
1 — cos b
cot 0 =
sin b
sin b
cot 0,
1 -f cos b
.
1 — cos a
,
cos (0—- 'x)
cos (C — x) = :
cot 0: therefore
sin a cos x
(1 -f- cos b) (1 — cos a)
or, cos C + sin C tan x —
sin a sin b
Substituting, in this equation, the values of cos C and sin 0, in
terms of the sides a, b, c, and putting for the sake of abridg¬
ment^ M = ^(l—cos2 a—cos2 b—cos2 c-|-2 cos a cos b cos c),
.
l+cosS—cose —cos a ,. , ,
we have tan x =
—
which determines
JV1
the angle ACI. From the isosceles triangles ACI, ABI,
B 0 I, we have ACI = ^ (C + A — B); and in the same
manner, B CI = £ (B + C — A); B AI = £ (A + B — C).
From which results these remarkable formulse,
* i, (A
, . +. C — TJX
1 + COS b — cos a — cos c
tan
B) ^^
60
SPHERICAL TRIGONOMETRY.
1 -j- cos a — cos b — cos c
tan J (B + 0 — A) =
^
1 + cos c — cos a — cos b
tan ^ (A + B — 0) = —^^
To which we may add that which gives cot S, and which
can he put under the form
„ , 4 +^
tani(A
B + C) = — 1 — cos a — cos b — cos c
From the value of the tangent of x, already found, we have
1 + tan2 x — -
1
2 (1 + cos b) (1 — cos c) (1 — cos a)
c M2
16 cos2
b sin2 J c sin2 J a
3
M2
4c cos J b sin -J- c sin J a
—
; but, from the equation
COS X J\x
1 — cos b
cos x = :—=— cot (b = tan i b cot g6, we find
sm 6
,
tan & b
,
4: sin i a sin ^ b sin i c
t
tan 0 = — ;
tan
0=
—^cos x
M
2 sin a sin % b sin % c
/( . a+6+c . a+b—c . a-f-c—ft- . 64-c —a\
8in
V V8'n
^— sm —2~sm
4:9. The surface of a spherical polygon is measured by the
sum of the angles, minus the product of two right angles, and
the number of sides of the polygon, minus 2.
From A draw the arcs AO and AD, the angles of the poly¬
gon ; it will then be divided into as many triangles, minus
two, as the figure has sides; but the surface of each triangle
is measured by the sum of the angles, minus two right
angles; and it is clear, that the sum of all the angles of the
triangles is equal to the sum of all the angles of the polygon.
i
SPHERICAL TKIGONOMETRY.
61
Therefore the surface of the polygon
is equal to the sum of the angles
diminished by as many times two
right angles as the figure has sides,
minus two.
Thus if S = sum of the angles
of a spherical polygon, n = the
number of its sides, then the surface
of the polygon is S — wtt + 2 7r =
S — 2 (n — 2) or S — 2 n -f- 4, when
the right angle is taken equal to unity.
POLYHEDRONS.
50. If S be the number of solid angles of a polyhedron, H
the number of faces, A the number of its edges, then
S + H = A + 2.
Take a point within the polyhedron, and from which draw
lines to all the angular points of the polyhedron; imagine
from this point, as a centre, we describe a spherical surface
which meets all these lines in as many points, then join these
points by arcs of great circles, in such a manner as to form,
upon the surface of the sphere, the same number of polygons
as there are faces of the polyhedron.
Let ABODE be one of these polygons, and n the number
of its sides, its surface by the last article will be S — 2 n + 4;
S being the sum of the angles A, B, 0, D, E. Similarly if we
find the value of each of the other spherical polygons, and
add them all together, we conclude that their sum or the
surface of the sphere which is represented by 8, is equal to
the sum of all the angles of the polygons, less twice the
number of their sides, plus four times the number of faces.
Now as the sum of all the angles that meet at the same point,
A, is equal to four right angles, the sum of all the angles of the
polygons is equal to four times the number of solid angles,
it is therefore equal to 48. Then double the number of sides
AB, BC, CD, &c., is equal to four times the number of edges,
or equal to 4 A, since the same edge serves for two faces;
8 = 4S-4A + 4H; or
2 = S—A + H; orS + H = A + 2.
[,See Legendre's Geometry, pp. 228, 229."]
SPHERICAL TRIGONOMETRY.
Got. It follows that the sum of all the plane angles which
form the solid angles of a regular polyhedron, is equal to
as many times four right angles as there are units in S — 2,
S being the number of solid angles of the polyhedron.
The plane angles = sum of all the interior angles of each
face, which Prop. 32, of the 1st book of Euclid
i
= H (n — 2). tt
= 2 (A — H) tt (since n H = 2 A)
•= (S — 2) 2 tt (since A — H = S — 2).
51. There can be only five regular polyhedrons.
Since every face has n plane angles, the number of plane
angles which compose all the solid angles = wH = Sm = 2Ar
and by the last article S + H = A + 2 ;
H = — S, and A =
; substituting these,
S + »S-=S+2,
n 2
2nS + 2mS = m?iS -f2 n S + 2 m S — m nS = 4/1,
S|2(7i + m) — m n} =
g_
^71
2 (n + m) — m n
Now this must be a positive whole number, and in order
that it may be so 2(m + n) must be greater than mn; and,
therefore, — + — >
or — > J - —; but m cannot be
Tfl
7b 7b TTL
less than 3, therefore —
cannot be so small
as A — 4-3) or 4
ffl
J
consequently, since n must be an integer and cannot be less
than 3, it can only be 3, 4, or 5. In the same manner m
cannot be less than 3, therefore the values of m can only be
3, 4, or 5.
SPHERICAL TRIGONOMETRY.
63
52. To find the inclination of two adjacent a
faces of a polyhedron to each other.
Let A B be the edge common to the two
adjacent faces of a polyhedron, C and E the c
centres of the faces. Draw CO, EO, per¬
pendicular to the faces meeting each other
in 0; and CD and ED, perpendicular to
AB, the intersection of the planes ABC,
ABE, then the angle 0DE is the required
inclination.
Let n be the number of sides in each
face, m the number of plane angles in each
solid angle if from the* centre 0, and
radius equal to unity, describe a spherical triangle meeting
the lines OA, 0 0, 0 D in p, q, r, we shall have spherical
triangle p q r, in which we have the angle r a right angle,
the angle p =
and the angle q = —
m
n
cos p
and by right-angled triangles, cos qr = —- ;
but cos qr = cos COD = sin CDO =sin J C, C being the
angle CDE; then
TT
cos —
m
•in
sin
| 0=
TT
sm —
n
This equation is general, and applies successively to the
five polyhedrons, by substituting the values of m and n in
each case.
Tetrahedron m = 3, n =. 3. Hexahedron m — 3, n = 4.
Octahedron m = 4, n ass 3. Dodecahedron m = 3, n = 5.
Icosahedron m = 5, n = 3.
From the triangle pqr from which we have deduced the
inclination of the two adjacent faces, we have
CO TT
TT
cos pq~ cot p cot q2 : or ——- = cot — cot — :
OA m
n
64
^ SPHERICAL TRIGONOMETRY.
therefore, if we call R the radius of the sphere which cir¬
cumscribes the polyhedron, and r the radius of the sphere
inscribed in it, we shall have — = tan
tan —:
r m
n
la
and, by making the side AB = a, we have 0 A = —^— ;
TT
sin —
n
JL a2
2
2
and, consequently, R = r H—-—
sin2—
n
These two equations give for each polyhedron, the values
of the radii R and r for the circumscribed and inscribed
sphere. We have, supposing 0 known,
r = i-2 a cot — tan \2
0 and R2
= ir tan — tan ^2 0.
n
m
R
In the dodecahedron and icosahedron, — has the same
r
TT TT
value for both ; viz., tan — tan —. Therefore, if R be the
o O
same for both, r will also be the same; that is to say, if
these two solids are inscribed in the same sphere, they will
also circumscribe the same sphere, and vice versd.
The "same property holds with regard to the hexahedron
R
and octahedron, since the value of — is the same for one as
r
1
n*
X
T
the other; viz., tan — tan -j- •
53. To find the inclination of two adjacent faces in the five
regular polyhedrons.*
TT
cos
From the equation sin J 0 = TT, taking the tetrahedron
sin —
n
* Legendre, at page 312, of his Geometry, finds the inclination from
cos c — cos a cos b . 0
equation
cos C = : sin:—;
*
a sin b J see equation 3, page 5.
SPHERICAL TRIGONOMETRY. 65
where m = o and n = 3,
.
cos 60°
1 „
r
sm ^ 0 = ——^5 «= —7=-; /. cos 0 = 4.
3
sin 60°
x/ 3 '
In tlie hexahedron m = 3 and n = 4,
sin J 0 = C?S
= —7=r, and cos 0 = 0; /. 0 = 90.
^
sm 45
x/ 2
In the octahedron m = 4 and n = 3,
. , ^ .cos 45
, /2
^
^
smf 0 = -—ttt = a/
and cos 0 == — 4
^
sm 60
V 3
*
In the dodecahedron m = 3, n == 5,
•
cos 60
2
— y5
sin
%■ G = ———
=— —
, and cos 1
O = 7=sm 36
yio — 2 a/ 5 5 —V 5
In the icosahedron m = 5, n = 3,
. , ^
cos 36
l + '/T
„
ST
sm J C = . gA = 7=-, and3 cos C=
r— •
^
sm 60
2v 5
3
54. To find the solid content ofregular polyhedron.
The area of each face = ^ a2cot-; hence the area of the
4
n
Tl2 TT
surface of the polyhedron = H. - a cot and the solid conarea of the surface x by the altitude
_ . .
tent =
= 43 of the area
3
of the surface x by radius of inscribed sphere.
.
tt nra2 H
k
TT n 2
= 4-3 X II.a cot — x r = ——— cot—
4
n
12
n
^or since r = J a cot ^ . tan
= !i
^C0t2^tan^c
(1)
0^
(2)
66
SPHERICAL TRIGONOMETRY.
From either of these equations we can find the solid
content.
We shall here use the first. Taking the equations
H. tt
TT cfi
—
= tan
—/V*>
tan and R2 — r2
r
/V*
/M
m
n
, ^
4 sm2 —
n
we can find r and R.
In the tetrahedron m = 3, n =. 2>;
R
—
= tan 60 . tan 60 = \/ 3 x
T
(3 }•)2 — r2 =
3=3:7
R == 3 r.
or 8 r2 = y;
,
a2 a R
• • ^ = ht.
• Also
——r
24:' »' 2?=■
v/6
3
9
3
2^6
In the hexahedron m = 3, n = 4,
—
= tan 60 tan 45 =
r
v
3 -: r = 2
^R 2
= a
.
In the octahedron,m—.4, andn=3, andr=—R——.
V 6 v/2
In the dodecahedron, m — 3, n = 5,
r = 1^250 +110-v/s"; R = |(v/15+ \/3).
In the icosahedron, m — 5, n = S,
r = ^42+ 18 ST-, R = | \/ioTvV
SPHERICAL TRIGONOMETRY.
67
These being substituted in equation (1), page 65, we find
the solidity of each of the five polyhedrons.
a3 /—
For the tetrahedron, the solidity is — v 2;
a3.
For the hexahedron
For the octahedron
For the dodecahedron
For the icosahedron
o
v2.
\/470 + 210 v/5
—
v 14 + 6 v 5.
Examples.
1. Prove that the sine of the sun's declination (S) is a
mean proportional between the sines of its altitude at six
o'clock (A) and its altitude when due east "(A').
The circles being projected on the
meridian, S'M=altitude at six o'clock
= A; SC = altitude when due east,
cos S'Z = cos Z P . cos P S', or sin
A = sin lat. sin b; similarly cos S P =
cos SZ cos ZP, or sin 2 = sin A'sin
lat.; dividing one by the other we have
sin A
sin 2 or
———T.J
, sin A : sin S ::
sin S
sin A'
sin S : sin A'.
2. Find at what latitude the azimuth of a star, whose de¬
clination = 30°, is equal to 45° at the time of rising. (See
last figure.)
Let Z be zenith; P the pole; C H horizon; then cos P M =
cos P H cos M H, or sin d = sin lat. cos M Z P; but cos M Z P =
y=>
an< s n
l i ^
Latitude = 45°.
; .'.•£■ = sin lat. —, and sin lat. =
68
SPHERICAL TRIGONOMETRY.
3. Find the sun's zenith distance when due east on a day
whose declination (S) is given, in a given latitude (X). (See
fig. p. 67.)
Let z c be the prime vertical, P the pole, and S the sun;
then cos S P = cos S z cos z P, or sin B = cos S z sin. lat.j
,
~
sin $
whence cos S z = —
sin X
4. Find latitude of place from given declinations of two
stars, and the difference of their altitudes when they are on
the prime vertical.
Let S and 0 be two stars on the prime vertical; S C =
difference of altitudes, and from the three given sides of the
A S P 0 we can find the Z. 0, which, with C Po in the rightangled triangle zP C, will give us 2 P = 90 -latitude, and
the latitude is readily found.
5. Find the relation of the polar to the equatorial diameter
of the earth from the horizontal parallax of the moon accu¬
rately observed dt the same time in different latitudes.
Let 0 be the latitude, r == corresponding semidiameter of
earth, a = equatorial diameter, b = polar distance. Then
the equation to the ellipse is a2 y2 + b2 x2 = a2 b2; putting
a2 b 2
2
r cos 0 for x, and r sin 0 for y,
r— •
a r = 20 . 20 . , 7<>2
. a sin 0 + b cos2 0
Also, if 0' be another latitude, and / the radius corresponding,
r'2 =
d2 b2
. 2
,
Y7D •
a sm 0 + 6722 cos 2 0'
2
dividin
S one
the other,
_
_.
a r2
m2 sin2 0' + cos2 0'
and making m = -, we have ^ = -2 ^ +
and m =
/cos2 0'- r'cos^)
yf \ t2 sin2 0 — r 2 sin2 0 J
6. To find the augmentation, by parallax, of the moon's
diameter.
SPHERICAL TRIGONOMETRY.
69
Let M be the moon, o the place
of observation on the surface of the
earth, whose centre is 0; D = ap¬
parent diameter seen from 0, and
D' the apparent diameter seen from
0; then, since the body is the
same, D' : D :: CM : oM :: sin z' :
sin (z' — p) ;
D' =
sin z
D
sin (z' — p) '
.-.D'-D = D
sin z' — sin (z' —p)
sin (z' —p)
sin-^-p cos (z'— ^p)
augmentation sought.
sin {z! — p)
Note.—If A be the value of D' in the last, when the moon is on the
horizon, in which case z7 = 90°, and p becomes the horizontal parallax P;
z'r =^
, we.have D
ta =
D' = D sin
—
D . sin 90° r = D ; also
sm (r —p) sin (90 — p) cos r
A cos P.
7. From two he¬
liocentric longitudes
(AII and AS) and
latitudes (PR and
QS) to find the place
of the node (N) and
the inclination of the
orbit.
We have by Napier's " Circular Parts" (pages 11 and 12),
sin NR (or AR—AN)
' sinNS (or AS—AN) _
cotan PNR =
tan PR tan QS
sin AR cos AN — cos AR sin AN
tan PR
sin AS cos AN — cos AS sin AN
; hence,
tan QS
sin AN
sin AR .tan QS — sin AS sin PR
tan AN =
cos A N ' cos A R . tan Q S — cos A S tan P R'
2D
70
SPHERICAL TRIGONOMETRY.
AN = required longitude or place of the node; this being
found, we have again by Napier's analogies cot P N It (incli^ =
sin AS-AN) •
nation offi
the orbit)
tan Qo
8. Explain Flamstead's method of finding the sun's right
ascension.
Let b and o be the vernal and autumnal equinoxes.
m n When the sun,
—-j ] —between b and o,
S passing in the
ecliptic b m n o,
has equal declinations (ms and nv), its distances from the
equinoxes (sb and ov) are equal also. We can find when
the sun has equal declinations by observing zenith distances
for two or three days soon after the vernal equinox, and for
two or three days about the same distance of time before the
autumnal equinox, and then, by proportion, ascertain the
precise time when the sun's declinations are equal; also we
can find by proportion the differences of the right ascension
of the sun.and some star by observing the differences at noon'
for some days. Let E = bs — right ascension of sun after
vernal equinox; then 180° — E = b v = right ascension
when it has equal declination. A — right ascension of star
in former case, and A + P = right ascension of star in
latter case. Then we obtain by observation A — E and
(180° — E) — (A + P). Let these differences of right as¬
censions be D and D', that is, A — E = D; (180° — E) —
(A + P) = D'. Adding these, we have 180° — 2 E — P =
+ I)/
) — P; whence A = D+E
Z
is known; the value of p, or the change in right ascension
of a star between the times of observation, is given by
tables.
D +D'; E =
18QO
~(D
9. In finding the longitude at sea by the lunar method, to
correct the observed distance of the moon from a fixed star
from parallax and refraction.
This is a useful application of the case 1 of oblique angled
spherical triangles. (See page 18.)
71
SPHERICAL TRIGONOMETKY.
Let HP be the apparent distance,
R being the'Star, and P the moon.
As the star's position is elevated by
refraction, take IIS = refraction of
the star, and take
= moon's paJ
\
rallax — refraction, as the parallax
being greater than the refraction, the p
moon is depressed. Let Z be the
zenith; SM will be the true distance, and is thus found.
H = apparent altitude of the star, H' = apparent altitude
of moon, A = true place of star, A' = that of the moon,
A = difference of altitudes, a
difference of true altitudes.
Then in the triangle Z lip we have
cos Hp — cos Z R cos Zp
COS • r/ T) • r/ >
Sin Z It Bin Zap
1 — cos Z —
sin (Z p — Z R) — cos R,p cos A — cos R p
sin Z R sin Zp cos H cos H'
similarly,
1 — cos Z : ^
sin (Z S — Z M) — cos S M
cos a — cos S M
sin Z S sin Z M cos h cos h'
cos a — cos S M
cos A — cos R p
7 T<— = Tv
TTT" >
cos h cos h cos H cos H
an( we
i
have
cos.AcosA' ,
^ v
cos a — cos S M = — — (cos A — cos R p);
cos H cos H'
■r-"
therefore SM (the true distance)
cos a
cos A cos h' ,
— — (cos A — cos R»),
cos H cos H' v
knowing, by tables, the time when the moon was at this
observed distance, so corrected, from a known star, the time
it Greenwich is found.
10. To find when that part of the equation of time which
arises from the obliquity of the ecliptic is a maximum.
PHERICAL TRIGONOMETRY.
S = place of the sun; o S = I = longitude; o R = right
ascension = r; S R = a =
declination; /_ S o R = w =
obliquity of ecliptic. Then
cos w = cot I tan x (p. 12)
tan r
1 — cos w
tan v ■•1 + COS U)
tan I — tan r
sm(l—r) „
,
7—— = ——7; : ; therefore I — r (the equation of
tan I + tan r
sin (I + r)
,
time) is a maximum when sin (? + r) is a maximum, i. e.
when I + r = 90°.
Note.—"When this happens, tan I = cot r, and cot I = tan r, cos w =
tan2 r; and cot21 = cos w.
11. In the oblique-angled spherical triangle ABC. Given
the side AB 73° 13', the side BO 62° 42', the side AC 119° 5',
required the angles.
( A = 44° 18'
Ans. B =136° 40'
( 0 = 48° 48'
12. The latitudes and longitudes of three places on the
earth's surface, suppose London, Moscow, Constantinople,
being given as below : required the latitude and longitude of
that place which is equidistant from the former three ?
The latitude of London is 51° 30', the latitude and lon¬
gitude of Moscow 55° 45', and 38°, and those of Constan¬
tinople 41° 30' and 29° 15' respectively.
13. Given the latitude of three places, Moscow 55° 30',
Vienna 48° 12', Gibraltar 35° 30', all lying directly in the same
arc of a great circle. The difference of longitude between
Vienna (situated in the middle), and Moscow, easterly, is
equal to that between Vienna and Gibraltar, westerly. It is
required to find the true bearing and distance of each place
from the other, and the difference of longitude, according to
the convexity of the globe.
14. Four given equal spheres being placed in close contact
with each other, it is required to find the volume of the space
inclosed between them and the three triangular planes through
each three centres.
SPHERICAL TRIGONOMETRY.
73
15. A point P being taken in the surface of a sphere, let
a, /3 denote its spherical distances from two given points ;
then if m cos a + n cos = a constant quantity, m and n
being any given numbers, the locus of P will be a circle.
16. The base of a spherical triangle is given, and the sum
of the cosines of the angles at the base, to trace the locus of
its vertex.
17. The sides of a spherical triangle are produced to meet
again in three more points, thus forming, with the original,
four spherical triangles, which constitute Davies's "Associated
Triangles" (12th edit., Hutton's Course, vol.ii. p.41), rr^rz
are the radii of the inscribed, and R Ri R2 Rs the radii of the
circumscribed circles. Prove that
tan2 R + tan2~Ri + tan2 R2 + tan2 R3 =
cot2 r + cot2 ri + cot2 T2 + cot2 7*3.
18. A person engages to travel from London to Constan¬
tinople, and to touch the equator in his journey; required
the point of contact and the length of his track, admitting it
to be the shortest possible, and the earth a sphere.
19. The angular points of two triangular pyramids being
respectively situated on four converging lines in space, let the
corresponding faces be produced to meet; then will the four
lines of section be all situated in the same plane.
20. Given the longitudes- of two places, 6° 49', and 54° 00',
their respective latitudes 48° 23' 14", and 4° 56' 13'; find
their distance^ the longitudes being both west, and their lati¬
tudes both north.
INDEX.
Adams's method of reducing Napier's
analogies, 9.
Analogies, Napier's formulae, 8; Adams's
method of reducing, 9.
Angle, to reduce an, to the horizon, 26.
Angles and sides of spherical triangles,
relations between, 5.
Circle, arc of a, defined, 1.
Circle, great, of a sphere, defined, 1;
poles of a great circle, 1.
Circular parts, Napier's, described, H;
application to an ambiguous case, 15;
rules, 11; examples, 12.
Cosines, to determine the cos and sines
of a spherical triangle, in terms of the
cosines and sines of the sides, 3.
Definitions, 1.
Dodecahedron, to find the solid content,
66; solid content, 67.
Examples and problems, 67.
Fundamental formulae, 4.
Girard's theorem for the areas of spherical
triangles, 51.
Great circle, definition, 1.
Hexahedron, to find the solid content, 66;
solid content, 67.
Horizon, to reduce an angle to the, 26.
Icosahedron, to find the solid content, 66;
solid content, 67
Isosceles triangle, 3.
Legendre's theorem, for the solution of
spherical triangles whose sides are very
small compared with the radius of the
sphere, 52.
Lexell's theorem, explained, 57.
Lhuillier's formula for the area of a sphe¬
rical triangle, 50.
Napier's analogies: formulae, 8; Adams's
method of reducing, 9.
Napier's circular parts, described, 11;
rules, 11; examples, 12; application to
an ambiguous case, 15.
Oblique-angled spherical triangles, solu¬
tion of, 18, 35:—(1) given the sides, fo
find three angles, 18, 35; (2) given the
two sides, and the angle opposite to one
of them, to find the remainder, 19, 36;
(3) given the two sides, and the in¬
cluded angle, to find the remaining
angles and side, 20, 37; (4) given the
two angles, and the adjacent side, to
find the remaining angle and sides, 20,
39; (5) given two angles, and the side
opposite to one of them, to find the rest,
21, 41; (6) given the three angles, to
find the three sides, 22, 42.
Octahedron, to find the solid content, 66;
solid content, 67.
Parallelepiped, given the three edges and
angles between them, to find the solidity,
55; the same being given, to find the
diagonal, 56.
Polar triangle described, 2.
Polygon, spherical, to find the area, 60.
Polyhedrons, to find the inclination of two
adjacent faces in the five regular, 64.
Polyhedrons, solid content of regular, 65,
67; equations to, 63; to find the area
of formulae, 61; to find the inclination
of two adjacent faces to each other, 63.
Problems and examples, 67.
Quadrantal triangle, 3.
Quadrantal triangles, solution of, 33.
Eight-angled spherical triangles, 3; nu¬
merical solution of, 28.
Sides and angles of spherical triangles,
relations between the, 5.
Sines, to determine the sines and cosines
of a spherical triangle, in terms of the
sines and cosines of the sides, 3.
Solid, content of the five polyhedrons, to
find the, of the dodecahedron, 66; of
76
INDEX.
the hexahedron, 66; of the icosahedron, 66; of the octahedron, 66; of the
tetrahedron, 66.
Sphere, definition of the, 1; area of the
lune, is proportional to the angle, 44;
circumference, ratio of, to the surface,
44; every section of a sphere is acircle^
1; great circle of the sphere defined, 1;
to determine a line on the surface of a,
on which the vertices of all the trian¬
gles are situated. 57.
Spherical excess, defined, 45 ; formula
for, 46; remarks on, 45.
Spherical polygon, to find the area, 60.
Spherical triangle, defined, 1; angles of,
1; area of, described, 44; to find the
area in terms of the three sides, 48;
Girard's theorem, 51; Lhuillier's for¬
mula, 50; Legendre's theorem for the
solution of, whose sides are very small
compared with the radius of the sphere,
52; sum of a, 2; to determine the sines
and cosines of a, in terms of the sines
and cosines of the sides, 3; relation be¬
tween the sides and angles of, 5:—(1)
relation between three sides and an
angle, 5 ; (2) relation between two
sides and their opposite angles, 5; (3)
relation between two sides and their
included angle, and the angle opposite
one of them, 6; (4) relation between
one of the sides and the three angles, *
ambiguous cases, 23; properties of, 27;
oblique angled, solution of, 18, 35, see
also Oblique-angled triangles; rightangled, solution of, 10; right-angled,
numerical solution of, 28:—(1) when
the hypotlienuse and one side are given,
28; (2) when two sides are given, 30;
(3) when a side and its opposite angle
are given, 31; (4) when a side and its
adjacent angle are given, 32.
Spherical trigonometry described. 3; fun¬
damental formula, 4.
Supplemental triangle, 2; sum of a, 2.
Table of results from ambiguous cases. 23.
Tetrahedron, to find the solid content of
a, 66 ; solid content, 67.
Triangle, polar, 2; sum of a, 2.
Triangles, spherical, defined, L; angles of,
1; described, 3; to determine the sines
and cosines of, 3; isosceles, 3; quadrantal, 3; right-angled, 3; ambiguous
cases, 23; Girard's theorem for the
areas of, 51; Legendre's theorem for
the solution of, 52; oblique-angled
spherical, solution of, 18, 35; quadrantal, solution of, 33, see also Spherical
triangles.
Trigonometry, spherical, described, 3;
fundamental formula, 4.
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