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, 36°50′, 21°24′ and 14°. What is the height of the tower? Ans. 53.96-feet. Ex. 3. Investigate General Roy's rule for the spherical excess, given in the scholium to prob. 8. Er. 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. Er. 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 265 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 ecliptic being 23° 27' 40'. Ex. 10. Required the longitude of the sun, when his right ascension and declination are 32° 46′ 52", and 13°13′27′′N respectively. See the theorems in the scholium to prob. 12. Ex. 11. The right ascension of the star a Ursæ majoris is 162° 50′ 34′′, and the declination 62° 50′ N: what are the longitude 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 mean proportionals between the last and first degrees of latitude. Er. 14. Demonstrate that the degrees of the terrestrial meridian, in receding from the equator towards the poles, are L 2 increased increased very nearly in the duplicate ratio of the sine of the latitude. Ex. 15. If p 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 with the meridian, then is d = Required a pm p+(m-p) sin2 a' demonstration of this theorem. CHAPTER VI. PRINCIPLES OF POLYGONOMETRY. THE theorems and problems in Polygonometry bear an in- · timate connection 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 1. 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 Petersburg, in Phil. Trans. vol. 65, p. 282: but they were first demonstrated by Dr. Hutton, in Phil. Trans. vol. 66, pa. 600. Let Let ABCDEF be a polygon: then will AF AB. COS A + BC. COS CBFA + CD COS CDA ÁF + DE. COS DEAF + EF. COS EF AAF*. For, drawing lines from the several angles, respectively parallel and perpendicular to AF; it will be Ab AB. COS BAF, B bc = BB = BC. COS CBB = BC. COS CB ^AF, 2 -de EE =DE. COS DEɛ = DE. COS DE AF, eF= EF . COS EFE EF. COS EFAAF. But AF bc + cd + de + CF Ab; and Ab, as expressed above, is in effect subtractive, because the cosine of the obtuse angle BAF is negative. Consequently, AF AC + cd+de+ eF = AB. COŚ BAF + BC. COS CBAAF + &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 Trigonometry 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 Respective Inclinations to that Base. Thus, in the preceding figure, cc=CB. sin CB^FA+BA.sin A; or Cc CD. sin CD ^AF + DE . sin DE^ AF + EF. sin F. This is evident from an inspection of the figure. . Cor. 1. In like manner Dd: DE. sin DE AAF + EF. sin F, or Dd = CB. sin CB^FA + BA. sin a CD. sin CD AAF. 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 ABC. sin (A+B) + CD. Sin (A+B+C)+DE. sin (A+B+C+D) + EF. Sin (A+B + C + D + E) = 0. * 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 CRAFA denotes the inclination of the line CB to FA. Here 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+C), their values sin A. cos B+ sin B. Cos A, sin A. cos (B+C) + sin (BC). cos A, &c (ch. ii equa. v), and recollecting that = sin tang (ch. ii p. 55), we shall have, COS sin A.(AB+BC.COSB+CD.Cos(B+c)+DE.COS (B+C+n)+&c) +COSA. (BC.sin B+CD.sin(+C)+DE.cos (B+C+D)+&c)=0; and thence finally, tan 180° - A, or tan BAF = BC sin B+ CD. sin (B + C) + DE . Sin (B + C + D) + EF. Sin (B+C+D+E) 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 B, are known, we have AB sin B ; BC AB. COS B in both which expressions, the second term of the denominator will become subtractive whenever the angle ABC is acute, or в 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 İnclude. 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 = b. cos a^b + c. c = a. cos a^c + b. E D d a cos ac + d. cos and, A cos b^c + d. cos c^d, a. cos ad + b. cos b^d + c. cos c^S. 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+ bò.cos bad+co.cos c^d). In like manner, c2 = a2 + b2 + d3 — 2(ab . cos a^b + a3⁄4. cosa^d+b♪.cos b^d). &c. &c. Or, d Or, since bcc, b^8 = c + D 180°, c^&D, we have a2 = b2 + c2 + 8-2(bc. cos c-bò. cos (c+D)+co. cos D), · c2 = a2 + b2 + ♪2 — 2(ab. cos B — b♪ . cos (A+B) + a§. cos A). &c. &c. { The same method applied to the pentagon ABCDE, will give bd. cos (c + D) + be . cos (C + D + E) ? g2 = b2 + c2 + d2 + €2 − 2 ) bc. cos c — + cd. cos D — ce. cos (D + E) + 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. 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 Exterior* Angles Contained 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 PBC. But twice the surface of the tri angle PAD is (Mens. of Planes pr. 2 rule 2) AP PD. sin P (ABBP). (DC + CP). sin P; and twice the surface of the triangle PBC is BP. PC . sin P: therefore their difference, or twice the area of the trapezium, is = (AB. DC + AB .CP + DC. BP). sin P. Now, in ▲ PBC, sin Psin B: BC: PC, whence PC = BC. sin B sin P BC. sin c sin P sin c: BC: PB, whence PB = Substituting these values of PB, PC, for them in the above equation, and observing that sin P sin (PEC 4 PCB) = sin sum of exterior angles B and C, there results at length, Twice surface? }={ 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) twice area ACD = + BC. DC. 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 |