3. The space would be improperly occupied in giving particular descriptions of these instruments, and the manner of adjusting them for use. These matters will be much better learnt from an examination of the instruments, and a few explanatory remarks from a judicious tutor. Nor shall we here embarrass the computation and diagrams, by showing how to allow for the height of the instrument, for that is a matter which requires only a single hint from the person who teaches the use of the instrument employed. So again, in reference to chains, tapes, and other contrivances for measuring lines, descriptions are suppressed. It ought, however, to be observed, that whenever a base, or distance between two stations, is measured on sloping ground, it must be reduced to the corresponding horizontal line, before it is employed in the general computation, if horizontal angles are taken at its extremities with a theodolite. This is in all cases easily effected: for, if the sloping or hypothenusal line be regarded as radius, the corresponding horizontal line will be the cosine of the inclination of the plane on which the sloping line was measured. It is, therefore, simply requisite to multiply the natural cosine of the angle of inclination, by the length of the line measured, to obtain the true horizontal line. Suppose that on a plane inclined to the horizon in an angle of 44 degrees, a distance of 400 yards were measured, what would be the corresponding horizontal distance? Multiply nat. cos 44° = ⚫9969173 400 Product gives hor. dist. = 398-7669200 Here the difference is not 1 yard, or not a 320th part of the measured line. As this is not a greater deviation from accuracy than will occur in the usual processes for measuring distances with a chain or a tape, the reduction from the sloping to the horizontal line, may, in common cases, be disregarded, except the an gle of inclination exceed 41° or 5°. 4. There are several simple methods of approximating to the heights of both accessible and inaccessible objects, by means of shadows, mirrors, unequal vertical staffs, &c. But as these depend solely upon the prin ciple of similar triangles, and do not require the theorems and formula of trigonometry, properly so called, they are not described here. The student may, without being detained by farther observations, proceed to the solution of the following problems. EXAMPLE I. In order to ascertain my distance from a tower, which was inaccessible by reason of an intervening river, I measured on a horizontal plane a base line of 600 yards, and at each end took the angle included between the other end and the tower, finding them to be 57° 35′ and 64° 51′ respectively. Required the distance of the tower from each end of the measured base. In the annexed figure are given AB = 600 C CAB 57° 35′ B CBA 64° 51′ conseq. c = 180° — (a + b) = 57° 34′. Logs Hence, sin c....57° 34'....9-9263507 So is sin A 57° 35′.... 9·9264310 To BC... 600.11 2.7782316 and, So is sin в 64° 51′.... 9.9567437 To Ac... 643·49.... 2·8085443. Remark. These distances might have been determined without the aid of an instrument to measure angles. Thus, suppose in the continuations of the res pective lines CA, CB, two distances, AD, BE were measured, each 120; and suppose, on measuring the distances BD, AE, the former is found to be 660, the latter 672 yards; from these measures the required angles may be determined. For, in the triangle ABE thus formed, we have (Euc. ii. 12) AE2 = AB2 + BE2 + 2EB. BP. Whence, by transBut BP is position and division, BP = AF2 AB2 BE2 2EB the cosine of the angle ABC, to the radius AB; so that, dividing the preceding expression by AB, we shall have the cosine of that angle to radius 1. will give the cosine of CAB. Thus, cos ABC = 425 A like process AE2- AB2 BE2 2EB. BA cos 64° 51' - 6002 240.600 - 1202 Hence, the angles ABC, and BAC, being determined, the distances are found as before. EXAMPLE II. In order to find the distance between two trees A and B, which could not be directly measured because of a pool which occupied much of the intermediate space, I measured the distance of each of them from a third object c, viz. AC = 588, BC = 672, and then at the point c took the angle ACB between the two trees = 55° 40′. Required their distance. This is an example to case 2 of plane triangles, in which two sides and the included angle are given. The work, therefore, is left to exercise the student: the answer is 593.8 EXAMPLE III. Wanting to know the distance between two inaccessible objects, which lay in a direct line from the bottom of a tower on whose top I stood, I took the angles of depression of the two objects, viz. of the most remote 25°, of the nearest 57°. What is the distance between them, the height of the tower being 120 feet? A The figure being constructed, as HAD = 25° 30', hence BAD = BAH In right angled ▲ ABC. In right angled ▲ ABD. Such is the way in which this problem is usually solved: it may, however, be done more easily and concisely, by means of the natural tangents. For, if AB be regarded as radius, BD and BC will be the tangents of the respective angles BAD, BAC, and CD the difference of those tangents. It is, therefore, equal to the product of the difference of the natural tangents of those angles into the height AB. EXAMPLE IV. From the top of a hill I observed two mile-stones on a horizontal road, which ran straight from its bottom, and took their respective angles of depression below the horizontal plane passing through the place of my eye; that of the nearer mile-stone was 14° 3', that of the farther was 3° 56'. Required the height of the hill. The figure being drawn, it will be found analogous to that in exam. 3, to which, therefore, we shall refer. There are given CD = 1760 yards, the distance between the two mile-stones; ADB HAD 3° 56′; ACB = HAC 14° 3'. This admits of three distinct modes of determination, as below. 1st Method. The angle ACB is equal to the sum of the angles CAD and CDA (Euc. i. 32): therefore CAD = ACB - ADB = 10° 7'. Then, in the triangle ACD, it will be, as sin CAD CD:: sin CDA : CA. And, in triangle ACB, it will be, as rad. or sin B: AC:: sin c: AB = 166.85; and so, if it be required, is sin CAB: CB = 666 75. According to this method the logarithmic process will require eight lines. 2d Method. Since CAD is the difference of ACB and ADC, we have, sin (CD): CD :: sin D: AC = sin (C-D) CD sin c sin D CD. sin D Also, rad: AC sin (C-D) But (ch. i. (= sin (C-D) = sin (C-D) sin c sin D cosec (c-D). Or, making the terms homogeneous (ch. iv. 17), The logarithmic formula is, therefore, this: |