OBLIQUE ANGLED TRIANGLES. 171c. The following are the algebraic values for the four quadrants: Now, from 172, 173, and 174, we find the cosines of the angles A, B, and C. 2 a c b2 + a3 177. Cos. C And by substituting s = † the sum of 2 ba Also, we find in terms of the tangent 190. We can find in terms of sine Radius of the inscribed circle in a triangle a) • (8 b) • (s — c) which is the same as given in sec. 48. S 191. Radius of the circumscribing circle R 4 {s • (3 — a) • (3 — b) • (8 — c)} . 192. By assuming D the distance between the centres of the inscribed and circumscribed circles, we have D2 = R22 R r, and D (R2 - 2 R r) = 193. Area of a quadrilateral figure inscribed in a circle is equal to { (§ — a) • (§ —— b) • (8 — c) • (s — d) }, where s is equal to the sum of the sides. (s 194. Sides are to one another as the Sines of their Opposite Angles. ac sine A: sine C. Having two Sides and their contained Angle given to Find the other Side and Angles. 203. Rule. The sum of the two sides is to their difference, as the tangent of half the sum of the opposite angles is to the tangent of half their difference; i. e., a + b: a b:: tan. † (A + B) : tan. † (A — B). Here a is assumed greater than b... the < A is greater than B.-E. I., 19. (See fig. 12.) Now, having half the difference and half the sum, we can find the greater and lesser angles of those required for half the sum, added to half the difference greater<, and half the difference taken from the half sum lesser <. When the Three Sides of the Triangle are given to Find the Angles. b is to the other 205. Rule. As twice the base or longest side A C two sides, so is the difference of these two sides to the distance of a perpendicular from the middle of the base; that is, 2 b: a + c :: a-c: DE. B E the line bisecting the base; Here B D is the perpendicular, and because B C =a is greater than A B c, C D is greater than A D; because A is greater than < C, the < ABD is less than < CBD; therefore, the area of the ◇ C D B is greater than ▲ AD B; consequently, the base C D is greater than A D. Let DE d; now the A B C is divided into two right angled triangles A B D and C B D, having two sides and an angle in each given to Cosine A may be found by sec. 175, and cosine C by sec. 177. 206. Example. Let the <A 40° (fig. 5), <B B C equal to 64 chains, to find the side A C. By sec. 194, sine 40°: 64 chains :: sine 50° ; A C. 50°, and the side Or thus: Log. sine 50° 9,884254 1,806180 In like manner, by the same section, A B may be found, because angles A and B together = 90° < C = 90°. 207. In the ABC (fig. 12), let the angle A = 40°, ang'e B = 60o, consequently, <C = 80. Let B C = 64, to find the side A C. Or thus: 55,42528 Sum Log, sine 11,743711 Diff. 1,935643 86,227 AC. Note. Here ar. comp. signifies arithmetical complement. It is log. sine 40° taken from 10 (see sec. 158 c), or it is the cosecant of 40°. Given Two Sides and the Contained Angle to Find the Other Parts. 208. Example. Let A C=120, B C = 80, and < A C B = 40°, to find the other side, A B, and angles A and B. By sec. 203, 120 + 80 : 120 — 80:: tan. 70°: tangent of the half difference between the angles B and A. i, e., 200: 40:: tan. 70°: tan. ✈ dif. B i. e., 5 : 1 :: 2,747477: 0,549495 · A. 28° 47'. 50,182 =cos. <BX B C. cos. <C 27° 59/ Consequently 50,182 ÷ 70 = 0,716885 =cos. <A 44° 12/ and 91,838 ÷ 104 = 0,88305 Having the angles A and C, the third angle at B is given. 104, log. 2,0170333, ar. comp. 7,9829667 b = 142,02, log. 2,1523495, ar. comp. 7,8476505 2)19,9738338 Cos. <C 13° 59′ 36′′/ = log. sine 9,986169 ... the angle A 27° 59/12/. In like manner, cos. <B may be found by sec. 176. The same results could be obtained by using the formulas in sections 184 and 188. HEIGHTS AND DISTANCES. 211. In chaining, the surveyor is supposed to have his chain daily corrected, or compared with his standard. He uses ten pointed arrows or pins of iron or steel, one of which has a ring two inches in diameter, on which the other nine are carried; the other nine have rings one inch in diameter. The rings ought to be soldered, and have red cloth sewed on them. He carries a small axe, and plumb bob and line, the bob having a long steel point, to be either stationary in the bob or screwed into it, thus enabling the surveyor to carry the point without danger of cutting his pocket. A plumb bob and line is indispensable in erecting poles and pickets; and in chaining over irregular surfaces, etc., he is to have steel shod poles, painted white and red, marked in feet from the top; flags in the shape of a right angled triangle, the longest side under; some flags red, and some white. For long distances, one of each to be put on the pole. For ranging lines, fine pickets or white washed laths are to be used set up so that the tops of them will be in a line. Where a pole has to be used as an observing station, and to which other lines are to be referred, it would be advisable to have it white-washed, and a white board nailed near the top of it. His field books will be numbered and paged, and have a copious index in each. In his office he will keep a general index to his surveys, and also an index to the various maps recorded in the records of the county in which he from time to time may practice. In his field book he keeps a movable blotting sheet, made by doubling a thin sheet of drawing paper, on which he pastes a sheet of blotting paper, by having a piece of tape, a little more than twice the length of the field book. The sheet may be moved from folio to folio. One end of the tape is made fast at the top edge and back, brought round on the outside, to be thence placed over the, blotting sheet to where it is brought twice over the tape on the outside, leaving about one inch projecting over the book. He has offset poles,one of ten links, decimally divided, and another of ten or six feet, similarly divided, mounted with copper or brass on the ends. One handle of the |