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then measure the distance AE. Now, by similar triangles, we have,
AE : AF :: AC AD,
in which all the terms are known except AD, which may, therefore, be found. The distance AD being laid off from A, the point D, at which the perpendicular CD meets AB, becomes known. If we wish the length of the perpendicular, we use the proportion,
in which all the terms are known, excepting CD: therefore, CD is determined.
III. To determine the horizontal distance from a given point to an inaccessible object.
37. Let A be an inaccessible object, and E the point from which the distance is to be measured.
At E, lay off the right angle AED, and measure in the direction ED, any convenient distance to D, and place a staff at D. Then measure from E, directly towards the object
A, a distance EB of any convenient D
length, and at B, lay off a line BC perpendicular to EA. Measure along the line BC, until a person at D shall range the forward staff on the line DA. Now, DF is known, being equal to the difference between the two measured lines DE and CB. Hence, by similar triangles,
DF: FC :: DE: EA,
in which proportion all the terms are known, except the fourth,
38. At the point E, lay off EB perpendicular to the line EA, and measure along it any convenient distance, as EB.
At B lay off the right angle EBD, and measure any distance in the direction BD. Let a person
at D align a staff on DA, while a second person at B aligns it on BE: the staff will thus be fixed at C. Then measure the distance BC.
The two triangles BCD and CAE being similar, we have,
BC BD :: CE: EA,
in which all the terms are known, except the fourth, which is, therefore, found.
39. Let B be the given point, and A the inaccessible object; it is required to find BA.
Measure any horizontal base-line,
as BC. Then, having placed staves at B and C, measure any convenient distances BD and CE, such that the points D, B, and A, shall be in the same right line, as also, the points E, C, and A; then measure the diagonal lines DC and EB.
Now, in the triangle BEC, the three sides are known, therefore, the angle ECB can be found. In the triangle CDB, the three sides are also known, therefore the angle CBD can be determined. These angles being respectively subtracted from 180°, the two angles ACB and ABC become known; and hence,
in the triangle ABC, we have two angles and the included side, to find the side BA.
is usually a portion of a longer line, so that the line CAD is well marked by stakes or pins, before AB is measured.
IV. To prolong a line beyond an obstacle.
41. Let OA be the line to be prolonged. Lay off OAB = 120°, or CAB = 60°. Measure AB, of such length as to permit BC to be measured without meeting the obstruction. Make ABC= 60°, and measure BC, equal to AB. If A be not in sight from C,
make the angle BCP equal to 120°, and resume the survey of the line. AC is equal to AB or BC.
NOTE. This method may be employed in the absence of any angular instruments, by constructing an equilateral triangle
the 90-link point, let two assistants draw out at the 30, and at the 60 point, until the three lines are straight.
V. To find the altitude of an object, when the distance to the vertical line passing through the top of it is known.
42. Let CD be the altitude required, and AC the known distance.
From A, measure on the line AC, any convenient distance AB, and place a staff vertically at B. Then placing the eye at A, sight to the
object D, and let the point, at which the line AD cuts the staff BE, be marked. Measure the distance BE on the staff; then, AB : BE :: AC : CD,
whence CD becomes known.
If the line AC cannot be measured, on account of intervening objects, it may be determined by calculation, as in the last problem, and then, having found the horizontal distance, the vertical line is readily determined, as before.
APPLICATIONS TO HEIGHTS AND DISTANCES.
I. To determine the horizontal distance to a point which is inaccessible by reason of an intervening river.
43. Let C be the point. Measure along the bank of the river a horizontal base-line AB, and select the stations A and B, in such a manner that each can be seen from the other,
and the point C from both of them.
Then measure the horizontal angles
CAB and CBA, with an instrument adapted to that purpose.
Let us suppose that we have measured AB = 600 yards; CAB = A = 57° 35′′, and CBA = B = 64° 51′.
sin C: sin B :: AB AC,
and applying logarithms, we have,
(a. c.) log sin C (57° 34′)
log sin B (64° 51′)
II. To determine the altitude of an inaccessible object above a given
44. Suppose D to be an inaccessible object, and BC the horizontal plane from which the altitude is to be measured: then, if we suppose ᎠᏟ to be a vertical line, it will represent the required distance.
Measure any horizontal base-line, as BA; and at the extremities B and A, measure the horizontal angles CBA and CAB. Measure, also, the angle of elevation DBC.