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 +2£b, BP. Whence, by trans position and division, BP ÁÉ2-AB2 BE2 But BP is 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. A like process will give the cosine of CAB. Thus, cos ABC = AE AB2 BE2 2EB. BA 6722 - 6002-1202 240,600 Hence, the angles ABC, and BAC, being détermined, the distances are found as before. EXAMPLE II. In order to find the distance between two trees A and E, 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 250, of the nearest 57°. What is the distance between them, the height of the tower being 120 feet? A The figure being constructed, as in the margin, AB = 120 feet, the altitude of the tower, and AH the horizontal line drawn through its B top; there are given, “HAD — 25° 30′, hence BAD — BAH — HAD = 64° 30′ HAC = 57° 0', hence BAC BAH — HAC Hence the following calculation. D 33° 0'. In right angled A ABD. In right angled ▲ ABC. To BC..77.929 1:5916986 To BD..251 585 2·4006851 “Conseq. BD - BC=173.656 feet, the distance required. 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 Áв. Thus, nat. tan 641° = 2:0965436 nat, tan 33° = 0.6494076 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. - ADB The angle ACB is equal to the sum of the angles CAD and CDA (Euc. i. 32): therefore CAD = ACB = 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 = 19 H) COSEC (CD) sin c sin D cosec (c – CD. sin D Also, rad: AC sin (C-D) CD sin c sin D Hence AB = CD sin (C-D) 1 = sin (C-D) D). But (ch. i. Or, making the terms homogeneous (ch. iv. 17), The logarithmic formula is, therefore, this: log AB = log CD + log sin c + log sin D + log cosec (C-D) 30 [in the index of the log]. ........ Thus, log CD The sum ...... 3° 56'.. 8.8362969 log cosec (C-D) 10° 7′..107553442 30, is log AB 166 857 feet 2.2223462 This method is, obviously, applicable to all similar examples. 3d Method. If AB were radius, CD would evidently be the differ ence of the tangents of BAD and BAC, or, of the cotan gents of BDA and BCA. Hence AB = Thus, nat cot D=14-543833 nat cot c 3.995922 CD cot Dcot c From the comparison of these three methods, 'it "will appear that the second ought to have the preference to the first: and that, considering the time employed in looking out the logarithms, the third method is preferable to the second. EXAMPLE V. Wanting to know the height of a church steeple, to the bottom of which I could not measure on account of a high wall between me and the church, I fixed upon two stations at the distance of 93 feet from each other, on a horizontal line from the bottom of the steeple, and at each of them took the angle of elevation of the top of "the steeple, that is, at the nearest station 55° 54', at the other 33° 20′. Required the height of the steeple. This is similar to exam. 4, and being worked by the 2d method, the height of the steeple is found to be 110.27 feet. EXAMPLE VI. Wishing to know the height of an obelisk standing at the top of a regularly sloping hill, I first measured from its bottom a distance of 36 feet, and there found the angle formed by the inclined plane and a line from the centre of the instrument to the top of the obelisk 41°; but after measuring on downward in the same sloping direction 54 feet farther, I found the angle formed in like manner to be only 23° 45'. What was the height of the obelisk, and what the angle made by the sloping ground with the horizon? The figure being constructed, as in the margin, there are given in the triangle ACB, all the angles and the side AB, to find вC. It will be obtained by this proportion, as sin c ( 17° 15′ = BA): AB (= 54) :: sin a (= 23°45′) : BC= 73.3392. Then, in the triangle DBC are known BC as above, BD = 36, 1 H B |