(4.) To find the distance between two points, one of which is accessible and the other Let P be the inaccessible point. A the accessible. inaccessible. Drive a picket in at A, and drive in another at any convenient point, B. Measure the angles PAB and PBA, also measure the line AB. Then by the fourth case of the solution of triangles (5). To find the distance between two points neither of which is accessible. Let A, B, be the two points; place on the A. ground two pickets C, and D, such that the distance between C and D can be measured, and that from each of C and D, the two inaccessible points and the other picket may be visible. Measure CD=p ACB C BCDC' D' Then in triangle ACD we know one side and two angles, and therefore can calculate AC. C D Fig. 5. In triangle BCD we know one side and two angles, and therefore can calculate BC. And finally in triangle ACB we have already calculated AC, BC, and have measured the angle C, and therefore can determine AB. Of course from the first two triangles we can also determine AD and BD, and then in triangle ABD we know two sides, and the included angle, and hence can determine AB, and can use these two calculations for checks upon each other. The calculation is performed as follows :— Call the sides and angles of ABC — a.b.c. A.B.C. and observe that CAD = 180°(C + C' + D') and CBD = 180° (C′ + D + D). Then from triangle ACD log. p + L. sin. (C + C' + D') + ar: comp: log. sin. D' Similarly log. a log. p + L. sin. (C' + D + D') + ar: comp: log. sin. C' - ACB we have A+ B C = 90° B tan. a + b A+ B tan. tan. (45°- ) (3) A+ B = L. tan. (45° whence we obtain B = Finally from triangle ABC and A and B since we know .. log. c = log. a + L sin. C + ar: comp: L. sin. A — 10 (5) It will be observed that in the above calculation we do not require a and b, but only log. a and log. b, which are given by equations (1) and (2). (Compare Plane Trig., Art. 40.) Let ABCD abcd be two rectangles. Then (Euclid. VI.-23.) the areas of these rectangles are to each other in the ratio com- = square foot, or a square yard, according as ab is an inch, a foot, or a yard. In this case Hence, if a and b be the sides of a rectangle, its area is ab, i.e., it contains as many units of area as the product of the number of units of length in one side, by the number of units of length in the other. Cor.-If a is the side of a square, its area is a2. If a.b be the sides of the parallelogram, and A the angle contained by those sides, then by the last article the area of half the parallelogram Or, if p is the perpendicular distance between two parallel sides, each of which a, then Or the area of a trapezoid sum of parallel sides + perpendicular distance between them. (5.) To find the Area of Trapezium. Let ABCD (Fig. 9), be the trapezium; join AC, from B and D let fall perpendiculars B m, Dn, upon A C. If instead of having the perpendiculars Bm and Dn given we have the sides AB. BC. CD. DA. and a A4 = S. (8. a) b) (s—c) Or, again, if we m Fig. 9. n have the sides and one angle, as A, given, we can proceed as follows. Let AB. a. BC= b. CD = c. DA= d. Join BD (Fig. 10), then we can calculate BD from triangle ABD by second case of oblique-angled triangles, and having calculated BD we C can determine the areas of the triangles as before. B A Fig. 10. (6). To determine the Area of an irregular Polygon. Suppose ABCDEF to be an irregular polygon, its area can be determined by effecting the following measurements; join AD, the longest distance across the figure from BCEF, draw perpendiculars to AD., viz. Bm, Cn, Ep, Fq, respectively; measure Am, mn, nD, Cn, Bm, Aq, qp, pD, Ep, Fq. Then area of polygon AB+ BinC + CnD + DpE + EpqF + FqA. = Am Bm + 1 mn × (Bm + Cn) + Cn x nD + · Dp + pE + 1⁄2 pq (Ep + Fq) + Aq × qF. (7.) To determine the Area of a regular Polygon. Various formulas have been already given for this in the "Treatise on Trigonometry" (Art. 45). It is there shown that if n be the number of sides of the polygon, and a the length of one of the sides, B m Fig. 11. 5 (√/5 + 1) a2 4 410-25 |