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FIGURE OF THE E ARTH.
fair in the Edinburgh Transactions, vol. v, to which, together
Having thus solved the chief problems connected with
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
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
Ex. 10. Required the longitude of the sun, when his right
Ex. 12. Given the measure of a degree on the meridian in
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 mean proportionals between the last and first degrees of
Ex. 14. Demonstrate that the degrees of the terrestrial meri-
Ex. 15 If Р be the measure of a degree of a great circle
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.
In Any polygon, any One Side is Equal to the Sum of all
* 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 ABCDEF be a polygon: then
For, drawing lines from the several angles, respectively parallel and perpendicular to AF; it will be
AB COS BAF,
BC. COS CBB
BC COS CBA AF, = CD. COS CDA AF, = DE . COS DEa AF,
EF . COS EFa AF
DE . COS DE:
EF. COS EFC
But AF = bc + cd + de + er
ab; and ab, as expressed
Cor. When the sides of the polygon are reduced to three,
The Perpendicular let fall from the Highest Point or Summit
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
Cor. 2. Hence the sum of the products of each side, into
perp. or 0. That is to say, in the preceding figure,
* 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.
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
ing that tang= (ch. ii p. 55), we shall have,
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 BCA =
AB. sin B
AB+BC. COS B
The Square of Any Side of a Polygon, is Equal to the Sum of
D C C
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.
cos b^c + d. cos bad,
a = b. cos a^b + c.
= a. cos a^b + c
c = α. cos a^c + b.
d a. cos aˆd + b . cos bad + c cos Cad.
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).
Or, since bac = c,
The same method applied to the pentagon ABCDE, will give
Cor. The property of a plane triangle expressed in equa. I
Twice the Surface of Any Polygon, is equal to the sum of
1. For a trapezium, or polygon of four
But twice the surface of the trian
gle PAD is (Mens. of Planes pr. 2 rule 2) AP. PD. sin P
sin P sin c :: BC: FB, whence PB =
Substituting these values of PB, PC, for them in the above
AB BC. sin B
+AB . DC. sin (B+C)
Cor. Since AB BC. sin B = twice triangle ABC, it follows
twice area ACD =
BC. Sin B
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.