FIGURE OF THE E ARTH.

fair in the Edinburgh Transactions, vol. v, to which, together
with the books mentioned at the end of the 1st section of this
chapter, the reader is referred for much useful information
on this highly interesting subject.

Having thus solved the chief problems connected with
Trigonometrical Surveying, the student is now presented with
the following examples by way of exercise.

Ex. 1. The angle subtended by two distant objects at a third object is 66° 30′ 39′′; one of those objects appeared un der an elevation of 25′ 47′′, the other under a depression of 1. Required the reduced horizontal angle. Ans. 60° 30′ 37′.

Ex. 2. Going along a straight and horizontal road which
passed by a tower. I wished to find its height, and for this
purpose measured two equal distances each of 84 feet, and at
the extremities of those distances took three angles of eleva-
tion of the top of the tower, viz. 36o 50′, 21° 24′, and 14o.
What is the height of the tower ?
Ans. 53.96 feet.

-95

#

Ex. 3. Investigate General Roy's rule for the spherical excess, given in the scholium to prob. 8.

Ex. 4. The three sides of a triangle measured on the earth's surface (and reduced to the level of the sea) are 17, 18, and 10 miles: what is the spherical excess ?

Ex. 5. The base and perpendicular of another triangle are 24 and 15 miles. Required the spherical excess.

Ex. 6. In a triangle two sides are 18 and 23 miles, and they include an angle of 58° 24′36′′. What is the spherical excess ? Ex. 7. The length of a base measured at an elevation of 38 feet above the level of the sea is 34286 feet: required the length when reduced to that level.

Ex. 8. Given the latitude of a place 48° 51′n the sun's declination 18° 30′N, and the sun's altitude at 10h 11m 26s AM, 52°35′; to find the angle that the vertical on which the sun is, makes with the meridian.

Ex. 9. When the sun's longitude is 29° 13′ 43′′, what is his
right ascension? The obliquity of the elliptic being 23° 27′ 40′′.

Ex. 10. Required the longitude of the sun, when his right
ascension and declination are 32o 46′ 52′′ and 13° 13′ 27′′. N
Voodik
respectively. See the theorems in the scholium to prob. 12.
Ex. 11. The right ascension of the star & Ursæ majoris is
162° 50′ 34′′, and the declination 62o 50'N: what are the longi-
tude and latitude? The obliquity of the ecliptic being as above.

Ex. 12. Given the measure of a degree on the meridian in
N. lat. 49°3′, 60833 fathoms, and of another in N. lat. 12°32′,
60494 fathoms: to find the ratio of the earth's axes.

Ex, 13. Demonstrate that, if the earth's figure be that of an oblate spheroid, a degree of the earth's equator is the first of

two

i

i

96

POLYGONOMETRY.

two mean proportionals between the last and first degrees of
latitude.

Ex. 14. Demonstrate that the degrees of the terrestrial meri-
dian, in receding from the equator towards the poles, are in-
creased very nearly in the duplicate ratio of the sine of the la-
titude.

❤

Ex. 15 If Р be the measure of a degree of a great circle
perpendicular to a meridian at a certain point, m that of the
corresponding degree on the meridian itself, and d the length
of a degree on an oblique arc, that arc making an angle a
pm
p+ (m-p) sin2 a Required
(m—p)

with the meridian, then is d=

a demonstration of this theorem.

PRINCIPLES OF POLYGONOMETRY.

THE theorems and problems in Polygonometry bear an intimate connexion and close analogy to those in plane trigonometry; and are in great measure deducible from the same common principles. Each comprises three general cases.

1. A triangle is determined by means of two sides and an angle; or, which amounts to the same, by its sides except one, and its angles except two. In like manner, any rectilinear polygon is determinable when all its sides except one, and all its angles except two, are known.

2. A triangle is determined by one side and two angles ; that is, by its sides except two, and all its angles. So likewise, any rectilinear figure is determinable when all its sides except two, and all its angles, are known.

3. A triangle is determinable by its three sides; that is when all its sides are known and all its angles, but three. In like manner, any rectilinear figure is determinable by means of all its sides, and all its angles except three.

In each of these cases, the three unknown quantities may be determined by means of three independent equations; the manner of deducing which may be easily explained, after the following theorems are duly understood.

THEOREM I,

In Any polygon, any One Side is Equal to the Sum of all
The Rectangles of Each of the Other Sides drawn into the
Cosine of the Angle made by that Side and the Proposed
Side*.

* This theorem and the following one, were announced by Mr. Lexel of Pe tersburg, in Phil. Trans. vol. 65, p. 282: but they were first demonstrated by Dr. Hutton, in Phil. Trans. vol 66, pa. 600,

Let

ས་ 1

Let ABCDEF be a polygon: then
will aF=AB. COS A+BC. COS CBa
FA + CD. COS CDA AF + DE COS
DEA AFEF. COS EFa AF*.

For, drawing lines from the several angles, respectively parallel and perpendicular to AF; it will be

Ab

bc

POLYGONOMÈTRY.

cd=

de

AB COS BAF,

BB

do

#E

BC. COS CBB
CD. cos CD

BC COS CBA AF, = CD. COS CDA AF, = DE . COS DEa AF,

EF

EF . COS EFa AF

DE . COS DE:

er

EF. COS EFC

But AF = bc + cd + de + er

ab; and ab, as expressed
above, is in effect subtractive, because the cosine of the ob-
tuse angle BAF is negative. Consequently,
AF=Ac+cd+de+er AB. COS BAF+BC. COS CBA AF+ &c. as
in the proposition. A like demonstration will apply, mutatis
mutandis, to any other polygon.

Cor. When the sides of the polygon are reduced to three,
this theorem becomes the same as the fundamental theorem
in chap. ii, from which the whole doctrine of Plane Trigono-
metry is made to flow.

THEOREM II.

The Perpendicular let fall from the Highest Point or Summit
of a Polygon, upon the Opposite Side or Base, is Equal to
the Sum of the Products of the Sides Comprised between
that Summit and the Base, into the Sines of their Respec-
tive Inclinations to that Base.

Thus, in the preceding figure, cc=CB. sin OB FA+BA sin 4; or ce=cD. sin CDAAFDE. Sin DEAAF+FF. Sin F. This is evident from an inspection of the figure.

Cor. 1. In like manner od-DE. Sin DEAAFEF. sin, or
pd=CB. sin CBAFA+BA. sin A-CD . sin CD^^F.

Cor. 2. Hence the sum of the products of each side, into
the sine of the sum of the exterior angles, (or into the sine of
the sum of the supplements of the interior angles), comprised
between those sides and a determinate side, is
+perp.

=

perp. or 0. That is to say, in the preceding figure,
AB. Sin A + BC. sin (A+B) +CD. sin (A+B+c)+DE. Sin
A + B + C + D) + EF. Sin (A + B + C + D + E) =0.

(A

* When a caret is put between two letters or pairs of letters denoting lines, the expression altogether denotes the angle which would be made by those two lines if they were produced till they met, thus c^ra denotes the inclination of

the line CB to FA.

VOL. 11.

14

Here

!

T

98

POLYGONOMETRY.

Here it is to be observed, that the sines of angles greater than 180° are negative (ch. ii equa. vII).

Cor. 3. Hence again, by putting for sin (A+B), sin (A + B +e), their values sin a. cos B + sin B. COS B+ sin B. cos a, sin ▲ . cos (E +c) + sin (B+c) cos A, &c. (ch. ii equa. v), and recollect

+33)

sin

ing that tang= (ch. ii p. 55), we shall have,

COS

sin ▲. (AB+BC. COS B+CD. COS (B+c)+DE. COS (B+C+D) + &c.) + cos A. (BC sin B+CD). sin (B+c)+DE. COS (B+C +D)+&c.)=0; and thence finally tan 180o-A, or tan BAF= BC. sin B+CD. sin (B+C) +DE. sin (B+C + D) + EF. Sin (B+C+ D + B) AB+BC.COS B+CD.Cos(B+c)+DE. COS (B+C+D) + EF. COS (B+C+D+E) A similar expression will manifestly apply to any polygon; and when the number of sides exceeds four, it is highly useful in practice.

Cor. 4. In a triangle ABC, where the sides AB, BC, and the angle ABC, or its supplement в, are known, we have

BC. Sin B

tan CAB

tan BCA =

AB. sin B
BC+AB. COS b

AB+BC. COS B
in both which expressions, the second term of the denomina-
tor will become subtractive whenever the angle ABC is acute,
or B obtuse.

THEOREM III.

The Square of Any Side of a Polygon, is Equal to the Sum of
the Squares of All the Other Sides, Minus Twice the Sum
of the Products of all the other Sides Multiplied two and
two, and by the Cosines of the Angles they Include.

D C C
Z

For the sake of brevity, let the sides be represented by the small letters which stand against them in the annexed figure : then, from theor. 1, we shall have the subjoined equations, viz.

A
cos a^c + d. cos aˆd;

cos b^c + d. cos bad,
cos bac + d. cos cad.

a = b. cos a^b + c.

b

= a. cos a^b + c

·

c = α. cos a^c + b.

a

d a. cos aˆd + b . cos bad + c cos Cad.

.....

a B

•

Multiplying the first of these equations by a, the second by b, the third by c, the fourth by ; subtracting the three latter products from the first, and transposing b2, c2, d2, there will result,

a2 = b2+c2+d2 -2(bc.cos b^c+bd. cos bad+cd. cos cd). In like manner,

c2 =α2 + b2 + d2

2(ab cos a+b+ad.cos ad+bd. cos b^d).
&c.

&c.

Or,

99

Or, since bac = c,
=c, bad c+p-180°, CAD, we have
a2 = b2 + c2 + 82 -2(bc, cos c—be . cos (c+D)+co. cost),
c2 = a2 + b2 + d2-2(ab. cos B-bo. cos (A+B)+að. cósa).

&c. &c.

+8)}}

2

The same method applied to the pentagon ABCDE, will give
a2 2=b2 +c2 +¿a fe3 bc. cos c-bd. cos (c+ D) + be. cos (c+D+E)
+ cd. cos D-ce. cos (DE) + de cos E
And a like process is obviously applicable to any number of
sides; whence the truth of the theorem is manifest.

Cor. The property of a plane triangle expressed in equa. I
ch. ii, is only a particular case of this general theorem.

THEOREM IV.

POLYGONOMETRY.

Twice the Surface of Any Polygon, is equal to the sum of
the Rectangles of its Sides, except one, taken two and two,
by the Sines of the Sums of the Extérior* Angles Contain-
ed by those sides.

1. For a trapezium, or polygon of four
sides. Let two of the sides AB, DC, be
produced till they meet at P. Then the
trapezium ABCD is manifestly equal to the
difference between the triangles PAD and P

D

PBC.

But twice the surface of the trian

gle PAD is (Mens. of Planes pr. 2 rule 2) AP. PD. sin P
(AB+BP). (DC+Cr). sin r; and twice the surface of the tri-
angle PBC is = BP . PC sin r: therefore their difference, or
twice the area of the trapezium is (AB. DC+AB. CP+DC.
BP) . sin r. Now, in ▲ FBC,

sin P sin c :: BC: FB, whence PB =

:

Twice surface

of trapezium.

sin P

Substituting these values of PB, PC, for them in the above
equation, and observing that sin P = sin (PBC+PCB)=sin sum
of exterior angles в and c, there results at length,

AB BC. sin B

+AB . DC. sin (B+C)
+BC. DC. sin c.

Cor. Since AB BC. sin B = twice triangle ABC, it follows
that twice triangle ACD is equal to the remaining two terms, viz.
AB. DC. sin (B+c)
+BC. DC.

twice area ACD =

sin c.

BC. Sin B
sin P

BC. sin c

* The exterior angles here meant, are those formed by producing the sides in the same manner as in th. 20 Geometry, and in cors. 1, 2, th. 2, of this chap.

2. For