(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 Λ. 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' ADB D ADC D' Then in triangle ACD we know one side and two angles, and therefore can calculate AC. 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 = (C + C' + D') and CBD = 180° (C′ + D + D′). Then from triangle ACD 180° or a tan. a tan. • a 10 log. 6 10 (1) Similarly log. a log. P + L. sin. (C' + D + D') + ar : comp: log. sin. C'— 10 (2) A + B с From triangle ACB we have 90° 2 6 A + B 2 6 1 assume tan. = tan. (45o – 0) 1 + tan. 0 (3) A +B (4) 2 A + B 2 sin. C C= a sin, A .. 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.) Example. Given CD 372.5 yds. C = 123°.15”. C = 13°.42. D = 129°.11. D' 19o.13'. (1) Log. b log. 372.5 2.5711263 9.6064647 2.6602086 2.5711263 9.4876426 2.6843172 = 2.6602086 + ar. comp. log. a 7.3156828 9.9758914 9.9757318 12654 33060 29526 35340 A + B (4) L. tan. L. tan. (45o – 0) + L. tan. 10. 2 2 (2) log. b L. tan. 1° 35' 22"2 = 8.4432492 8:1758516 8:1755658 1408)2858(2 2816 42 28° 22' 30" A 29° 14' 2" B 27° 30' 58" (5). Ieg. c = log. a + L. sin. C + ar. comp. L sin. A 10. = log. a 2:6843172 L. sin. 56° 45' 9.9223549 ar. comp. L. sin. 29° 14' 2" •3112458 2.9179179 82778 9179149 6 30 827.786 32 ..c= 827.786 yds. 10 (Answer.) II. THE MENSURATION OF AREAS. D с в (1.) To find the Area of a Rectangle. Let ABCD abcd be two rectangles. Then (Euclid. VI.—23.) the areas of these rectangles are to each other in the ratio compounded of the ratios of the sides, i.e., in the ratio compounded of the two, ab : AB and bc : : d BC, and if we suppose these lines to be represented by numbers, this compounded ratio is ab x bc: AB x BC. Hence rectangle ac : rectangle BC :: ab x bc : AB x BC. Fig. 6. Now suppose ab 1, and cb : 1. Then the area ac is the unit of area, i.e., is a square inch, or a square foot, or a square yard, according as ab is an inch, a foot, or a yard. In this case Rectangle BC AB x BC. Hence, if a and 6 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 a?. (2). To find the Area of a Triangle. А N Let ABC be the triangle; from A draw AN perpendicular to BC. Then, since the area of ABC is half that of a rectangle, whose base is BC and BC X AN height AN, the area of triangle 2 Cor.-(1). If we have given AB. BC and angle B. BC x AB sin. B ac sin. B B Area triangle = 2 2 sin. C sin. A a sin. B sin. C 2 sin. A where 28 = a + b + C. с a (3). To find the Area of a Parallelogram. 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 ab sin. A 2 :. Area of parallelogram = ab sin. A, Or, if p is the perpendicular distance between two parallel sides, each of which == a, theu Area of parallelogram = ap. (4.) To find the Area of a Trapezoid. D Let ABCD be the trapezoid of which the side AB is parallel to the side CD. Join BD, draw DM perpendicular to AB and BN perpendicular to DC, or DC produced, and let a = AB. b = DC and A p=DM or BN. Then Fig. 8. Area triangle ABD = de ap Area triangle BCD ļ bp .. Area ABCD = p (a + b) Or the area of a trapezoid = } sum of parallel sides + perpendicular distance between them. B M (5.) To find the Area of Trapezium. Let ABCD (Fig. 9), be the trapezium; join AC, om B and D let fall perpendiculars B m, Dn, of whi Or, again, if we have the sides and one angle, as A, given, we can proceed as ollows. 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 ›blique-angled triangles, and having calculated BD we can determine the areas of the triangles as before. B A Fig. 10. C (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 + BinnC + CnD + DpE + EpqF + FqA. (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, de C to A Ca B Fig. 11. n 5a3 Area |