= CBD41; to find the other angles, and the side CD.. Thus, first, as CB+ BD CB BD:: tan (D+ c) (1399): tan (DC) 42° 24'. Hence 69° 30′ +4224112° 54 CDB, and 69° 30' 42° 24 265 BCD. Then, sin BCD: BD :: sin CBD : CD =51.86, height of the obelisk. The angle of inclination DAEHDA =22° 54' CDB - 90° Remark If the line BD cannot be measured, then the angle DAE of the sloping ground must be taken, as well as the angles CAB, and CBD.. In that case DAE +90° will be equal to CDB: so that after CB is found from the triangle ACB, CD may be found in the triangle CBD, by means of the relation between sides and the sines of their opposite angles. EXAMPLE VII. Being on a horizontal plane, and wanting to ascertain the height of a tower standing on the top of an inaccessible hill, I took the angle of elevation of the top of the hill 40%, and of the top of the tower 51°, then measuring in a direct line 180 feet farther from the hill, I took in the same vertical plane the angle of elevation of the top of the tower 33° 45', Required from hence the height of the tower. The figure being constructed, as in the margin, there are given, AB 180, CAB 33° 45′ ACB CBE CAE 17° 15' CBD = 11, BDC 180° - (90° DBE) 130°. And CD may C be found by two proportions, viz. A200 BE 1st. As sin ACB: AB:: sin CAB: CB, and 2dly, as sin D : CB:: sin CBD: CD. This process would require eight lines. But the operation may be shortened; for, by the principles of method 2, exam. 4, we shall have CD rad4 AB sin A sin CBD cosec ACB sec DBE. At the top of a castle which stood on a hill near the sea-shore, the angle of depression of a ship at anchor was observed to be 4° 52′; at the bottom of the castle the angle of depression was 4° 2. Required the horizontal distance of the vessel, and the height of the hill on which the castle stands above the level of the sea, the castle itself being 60 feet high. H In the annexed diagram, where HT, OB, are parallel, and AT perpendicular to the horizontal line AS, are given BT = 60 feet, HTS 4° 52', consequently ATS = 85° 8', and OBS = 4° 2′, whence ABS exam. 4, is obviously applicable: so that we have 85° 58'; to find As and AB. AS. rad3 T Here method 2, BT sin ATS sin ABS COSEC (ABS - ATS) The logarithmic operation will stand thus: log TB.....60 1.7781513|log TB .....60 1.7781513 sin ATS 85° 8' 9-9984315 sin ATS 85° 8′ 9-9984315 sin ABS 85° 58′ 9-9989230 cos ABS 85°58′ 8.8471827 Cosec (B-T)50′11-8371392 cosec (в — T)50′ 11·8373192 log AS 4100-4 ft. 3.6128250 log AB 289-12 ft. 2·4610847 2 EXAMPLE IX. C Wanting to know the distance of an object at D from two others A and B in the same horizontal plane, as well as the distance between A and в, a pole was set up at c in a right line with AB, and the angle ACD was found to be 57°. The distance CD being measured was found to be 549.36 yards; and at D the angles CDA and ADB were taken; the first = 14°, the latter 41° 30'. Required the above specified distances. Here, the sum of the angles c and CDA taken from 180°, leaves the angle CAD, and the sum of the angles c and CDB taken from 180°, leaves the angle CBD. Then, it will be as sin CBD: CD:: sin C: DB. As sin CAD: CD :: sin C: DA. And sin ABD: AD :: sin ADB: AB. These operations will give DB = 498'68, da = 487·27, AB = 349.52 yards. EXAMPLE X. A B Wanting to know the horizontal distance between two inaccessible objects A and B, and not finding any station from which I could see them both, I chose two points c and D distant 200 yards, from the former of which A could be seen, from the latter B, and at each of the points c and d a flag-staff was set up. I then measured Fc 200, DE 200, and, hav- F C ing set up flag-staves at F and E, took E the following angles, viz. AFC = 83°, ACF = 54° 31′, ACD = 53° 30′, BDC = 156° 25′, bde = 54° 30′, and BED = 88° 30'. Required AB. Here, in the triangle AFC, all the angles are given, and the side FC, to find AC. Then, in the triangle ACD are given AC, CD, and the contained angle, to find the other angles and the side AD. Next, in the triangle BDE are given DE and all the angles to find DB. Lastly, in the triangle ADB are known AD, DB, the included angle BDA = BDC — ADC, to find AB = 345;5 yards. EXAMPLE XI. In order to determine the distance between two inaccessible objects E and w, on a horizontal plane, we mea sured a convenient base AB of 536 yards, and at the extremities A and B took the following angles, viz. BAW 40°16', WAE= 57°40′, ABE 42° 22′, EBW = 71°7′. Required the distance Ew. First, in the triangle ABE are given E all the angles, and the side AB, to find BE. So again, in the triangle ABW, are given all the angles and AB to find BW, Lastly, in the triangle BEW are given the two sides EB, BW, and the included angle EBW, to find EW = 939.52 yards, W Remark. In like manner the distances taken two and two, between any number of remote objects posited around a convenient station line, may be ascertained. EXAMPLE XII. B Suppose that in carrying on an extensive survey, the distance between two spires A and B has been found equal to 6594 yards, and that c and D are two eminences conveniently situated for extending the A bi triangles, but not admitting of the determination of their distance by actual admeasurement; to ascertain it, therefore, we took at c and d the following angles, viz. SACB = 85° 46′ LBCD = 23° 56′ ADC = 31°48′ | ADB = 68° · 2' D Required cp from these data. In order to solve this problem, construct, a similar quadrilateral Acdb, assuming cd equal to 1, 10, or any 1 other convenient number: compute Ab from the given angles, according to the method of the preceding example. Then, since the quadrilaterals Acdb, ACDB, are similar, it will be, as ab: cd:: AB: CD; and CD is found 4694 yards. EXAMPLE XIII. If the height of the mountain called the Pike of Teneriffe be 3 miles, and the angle formed at its top between the vertical or plumb line and a right line conceived to touch the earth in the horizon, or at the farthest visible point, be 87° 46′ 33′′; it is required from hence to determine the diameter of the earth, supposing it to be a perfect sphere. B E T Let c, in the marginal diagram, be the centre of the earth, the circle BTG a vertical section passing through the centre, AB the height of the Pike of Teneriffe, and AT the tangential line drawn to the visible horizon: let, also, вE, a tangent to the earth's surface at B, meet the other tangent AT in E. Then, in the triangle ABE, right angled at B, and having the angle BAE G =87° 46′ 33", it is, as rad: AB:: tan A: BE, and :: sec A: AE. But it is evident, from Euc. iii. 37, that BE = ET; therefore, AE + EBAE ET AT. In the triangle ATC right angled at T, we have, as rad : AT:: tan A: TC, the radius of the earth. The operation performed as above described occupies but small compass; it may, however, be shortened by means of ch. ii. prop. 4. For, since tan A+ sec a = tan (A+comp. A), we shall, by incorporating the proportions from which AE, BE, and CT are deduced, have CT. rad2 = AB tan (A + comp. a) tan a. Or, log CT = log AB + log tan (A + comp. A) + log tan A - 20, in the index. R |