= Also, DB AD c (one figure), or c AD (other figure). Hence, in ACD, AC, AD are known. Also, in CDB, CB, DB are known. .. A can be found. .. B can be found. Another method of solution is given in Art. 58. 1. Ex. 1, Art. 58. 3. Ex. 1, Art. 62. EXAMPLES. 2. Ex. 2, Art. 58. 4. Other Exs. in Arts. 58, 62. 5. Solve some of the problems in Art. 63 by means of right-angled triangles. 34 a. The area of a triangle in terms of the sides. (See Fig. 30.) From (1), (2), Case IV., Art. 34, Let then = (a + b + c) ( − a + b + c) (a − b+c)(a+b−c). 4 c2 a+b+c=2s; 2(sa) = a+b+c-2a=a+b+c. Similarly, 2 (sb) ab+c; 2 (sc) = a + b-c. .. Area ABC = 1⁄2 AB · CD = √s(s − a)(s — b)(s — c).* Ex. Find the areas of the triangles in Exs. Case IV., Art. 34. Check the results by finding the areas by the method of Art. 31. *This is sometimes known as Hero's Formula for the area of a triangle. It was discovered by Hero (or Heron) of Alexandria, who lived about 125 B.C., and placed engineering and land surveying on a scientific basis. 34 6. Distance and dip of the visible horizon. Let C be the centre of the earth, and let the radius be denoted by r. its height PL be denoted by h. Join P, C; draw PB from P to any point in the visible horizon; draw the horizontal line PH in the same plane with PC, PB. Then angle HPB is called the dip of the horizon. By geometry, PB2 angle PBC= 90°. PC2-CB2 = (r + h)2 — r2 = 2 rh + h2. Since h2 is very small compared with 2 rh, Taker 3960 mi., and let h be measured in feet. Then of P in miles. P H h B FIG. 30a. Hence, the distance of the horizon in miles is approximately equal to the square root of one and one-half times the height in feet. EXAMPLES. 1. A man whose eye is 6 ft. from the ground is standing on the seashore. How far distant is his horizon? 2. Find the greatest distance at which the lamp of a lighthouse can be the light being 80 ft. above the sea level. seen, 3. Find the height of the lamp of a lighthouse above the sea level when it begins to be seen at a distance of 12 mi. What 4. From the top of a cliff, 40 ft. above the sea level, the top of a steamer's funnel which is known to be 30 ft. above the water is just visible. is the distance of the steamer? 5. Find the distance and dip of the horizon at the top of a mountain 3000 ft. high? 6. Find the distance and dip of the horizon at the top of a mountain 21 mi. high. 34 c. Examples in the measurement of land. In order to find the area of a piece of ground, a surveyor measures distances and angles sufficient to provide data for the computation. An account of his method of doing this, and of his arrangement of the data and the results in a simple, clear, and convenient form, belongs to special works on surveying. This article merely gives some examples which can be solved without any knowledge of professional details. The various rules for finding the area, of a triangle and a trapezoid, are supposed to be known. In solving these problems, the student should make the plotting or mapping an important feature of his work. The Gunter's chain is generally used in measuring land. It is 4 rods or 66 feet in length, and is divided into 100 links. An acre = 10 square chains=4 roods=160 square rods or poles. The points of the compass have been explained in Art. 30. EXAMPLES. 1. A surveyor starting from a point A runs S. 70° E. 20 chains, thence N. 10° W. 20 chains, thence N. 70° W. 10 chains, thence S. 20° W. 17.32 chains D1 C1 IN D H C to the place of beginning. What is the area of the field which he has gone around? Make a plot or map of the field, namely, ABCD. Here, AB represents 20 chains, and the bearing of B from A is S. 70° E. BC represents 20 chains, and the bearing C from B is N. 10° W., and so on. Through the most westerly point of the field draw a north-and-south line. This line is called the meridian. In the case of each line measured, find the distance that one end of the line is east or west from the other end. This easting or westing is called the departure of the line. Also find the distance that one Bend of the line is north or south of the other end. This northing or southing is called the latitude of the line. For example, in Fig. 30 b, the departures of AB, BC, CD, DA, are B1B, BL, CH, DD1, respectively; the latitudes of the boundary lines are AB1, B1C1, C1D1, D1A, respectively. It should be observed (Art. 36) that the algebraic sum of the departures of the boundary lines is zero, and so also is the algebraic sum of their latitudes. The following formulas are easily deduced : B FIG. 30b. L Departure of a line = length of line x sine of the bearing; Latitude of a line = length of line x cosine of the bearing. By means of the departures, the meridian distance of a point (i.e. its distance from the north-and-south line) can be found. Thus the meridian distance of C is CC, and C1C D1D+ HC. = B1C1, C1C, C1D1, DiD can be computed. Hence in Fig. 30 b, AB1, B1B, Now area ABCD = trapezoid D1DCC1 + trapezoid C1CBB1 – triangle ADD1 — triangle ABB1. The areas in the second member can be computed; it will be found that area ABCD = 26 acres. NOTE. Sometimes the bearing and length of one of the lines enclosing the area is also required. These can be computed by means of the latitudes and departures of the given lines. The formulation of a simple rule for doing this is left as an exercise to the student. 2. In Ex. 1, deduce the length and bearing of DA from the lengths and bearings of AB, BC, CD. 3. A surveyor starts from A and runs 4 chains S. 45° E. to B, thence 5 chains E. to C, thence 6 chains N. 40° E. to D. Find the distance and bearing of A from D; also, the area of the field ABCD. Verify the results by going around the field in the reverse direction, and calculating the length and bearing of BA from the lengths and directions of AD, DC, CB. 4. A surveyor starts from one corner of a pentagonal field, and runs N. 25° E. 433 ft., thence N. 76° 55′ E. 191 ft., thence S. 6° 41′ W. 539 ft., thence S. 25° W. 40 ft., thence N. 65° W. 320 ft. Find the area of the field. Deduce the length and direction of one of the sides from the lengths and directions of the other four. 5. From a station within a hexagonal field the distances of each of its corners were measured, and also their bearings; required its plan and area, the distances in chains and the bearings of the corners being as follows: 7.08 N.E., 9.57 N. E., 7.83 N. W. by W., 8.25 S. W. by S., 4.06 S.S.E. 7° E., 5.89 E. by S. 31° E. 35. Summary. Chapter II. was concerned with defining and investigating certain ratios inseparably connected with (acute) angles, and attention was directed to the tables of these ratios and their logarithms. In Chap. III. it was shown how these definitions and tables can be used in finding parts of a rightangled triangle when certain parts are known. In Chap. IV. the knowledge gained in Chap. III. was employed in the solution of some of the many problems in which right-angled triangles appear. In Art. 34 it has been seen that this knowledge can serve for the solution of oblique triangles. It follows, then, that it can serve for the solution of problems in which oblique triangles appear, and, accordingly, for the solution of all problems involving the measurement of straight lines only. Consequently, the student is now able, without any additional knowledge of trigonometry, to solve the numerical problems in Chaps. VII., VIII. It is thus apparent that even a slight acquaintance with the ratios defined in Chap. II. has greatly increased the learner's ability to solve useful practical problems. Oblique triangles can sometimes be solved in a more elegant manner than that pointed out in Art. 34. In order to show this, further consideration of angles and the trigonometric ratios is necessary. Consequently, in Chap. V. some important additions are made to the idea of a straight line and the idea of an angle; the trigonometric ratios are defined in a more general way, namely, for all angles, instead of for acute angles only, and the principal relations of these ratios are deduced. Chapter VI. treats of the ratios of two angles in combination. While it is necessary to consider these matters before proceeding to the solution of oblique triangles given in Chap. VII., it should be said that the knowledge that will be gained in Chaps. V., VI., VII., is necessary and important for other purposes besides the solution of triangles. In fact, the latter is one of the least important of the results obtained in these chapters. N.B. Questions and exercises suitable for practice and review on the subject-matter of Chap. IV. will be found at page 184. |