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The sum — 30, is log AB 166.857 feet 2.2223462
This method is, obviously, applicable to all similat examples.
3d Method. If AB were radius, cD would evidently be the difference of the tangents of BAD and BAC, or, of the cotancio
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 prefer
able to the second.
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 1 1027 feet.
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 2
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 be found by two proportions, viz. * 1st. As sin AgB : 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 rad" = AB sin A sin CBD coSec ACB sec DBE.
The sum - 40, is log co .. 83.9983 feet ... 19242707
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. In the annexed diagram, where HT, op, 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 = 85° 58'; to find As and AB. Here method 2, exam. 4, is obviously applicable: so that we have As. rad} = BT sin ATs sin ABs cosec (ABs — Ats) AB. rada = BT sin ATs cos ABs cosec (ABs - ATs). The logarithmic operation will stand thus:
log As 4100.4 ft. 36128250 log AB 289:12 ft. 2:46.10847 ExAMPLE IX. * 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 B, 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 c1) A and Ape 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 cBoo. 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. .
Wanting to know the horizontal distance between two inaccessible objects A and B, and not finding any station from which I could see them A B 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 mea- D sured FC = 200, DE = 200, and, hav- FTö E ing set up flag-staves at F and E, took the following angles, viz. Afc = 83°, AcF = 54° 31', AcD = 53° 30', EDc = 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 AgD are given Ac, co, and the contained angle, to find the other angles and the side AD. Next, in the triangle