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47. To find the height of an accessible object, as AB.
Recede from the object AE along the line ED, to some convenient distance, which measure. With the quadrant take the angle ACB; then in the triangle ACB, there is given the side BC, and the angle ACB; hence the side AB may be found by B (38 Pl. Trig.) and therefore the height required.
(N. B. BE or CD represents the E height of the eye of the observer above the horizontal plane.)
Example. Let the horizontal distance ED be 200 feet, and the angle of elevation ACB=37° 35'; supposing the height of the eye to be 5 feet; required the height of the tower. By (38 Pl Trig.) As rad. : tang ACB::ED or CB : BA. Neg. log radius
iog tang ACB (37° 35') -- 9.886288
2.187318 Whence, adding 5 feet for the height of the observer's eye above the horizon, we have AE the height of the tower=158.92 feet.
48. To find the height of an inaccessible object, . CD.
At any two convenient stations A, B in the same vertical plane
1) with CD, observe the angles of elevation DAC, DBC, and mea. sure the distance AB or GE Then because the exterior angle DBC is equal to the two interior angles
BA BDA, DAB; if DAB be subtracted from DBC, the angle BDA will remain. Therefore in the triangle ADB we shall have the side AB, and the angles DAB, ADB, whence the side DB will be found by Case I. Plane Trig. And then in the right-angled triangle DBC, we shall have the angle DBC, and the side ĎB just found, whence DC may be determined, and consequently DF, the height required.
Example. In the above figure let the angle of elevation DBC be 48°; and at another station in the same vertical plane, but 200 feet farther off in the same direction, let the angle of elevation be 26° 45', the height of the eye being 5 feet; required the height of the tower.
First L ADB=48°—26° 45=219 15'.
= 9.653308 log AB (200)
2.501030 log BD=248.87
= 2.266177 Whence DF=189.57 feet, the height of the tower.
49. To find the distance of an inaccessible object. Let C be any inaccessible object, and
C A, B two points from which the distance of that object is to be found. Measure the distance AB, and observe the horizontal angles BAC, ABC. Then in the triangle ABC, we have given the base AB, and the A
two angles A and B, whence the sides AC, BC may be found by (39 Trig.)
Example. Let the angle A=86° 52', the angle B=78° 47', and the base AB=10110 feet; what is the distance between A and C?
The proportion in (39 Trig.) gives
As sin (C)* : AB::sin B : AC, or by logarithms arith. com. log sin C (14° 21') 10.605821 log sin B (78° 47')
9.991624 log AB (10110)
4.004751 log AC
4.602196 Wtence AC=40012.5=7 miles, 4 fur. 137 yds. 1} feet. The perpendicular distance CD may be found by the proportion rad. ; sin A:: AC : CD.
50. To find the distance between two objects, the distance of each from a third being known, and also the angle at that object which the required distance subtends.
Let A and B be the two objects, whose distance is required, and C the third object.
c Then in the triangle CAB, we have given the two sides CA, CB, and the included angle ACB ; whence the required distance AB may be found by Case III. Plane Tri
From the latter part of 41 Plane Trigonometry it appears that AB=VAC? + BC?-2 AC x BC cos C, and by this formula we may find AB without first determining the angles A and B. As we have not before exemplified this formula, we shall apply it to the solution of an example under this case.
Example In order to determine the distance between two houses at A and B, which could not be done by direct measurement, I ascertained the distance of each from a point C, and found AC=600 yards, BC=700 yards; at C I then took the angle ACB=569 12. From these data I require the distance AB?
Because sin (A+B)=sin(180-(A+Bl=sin C.
Here AC’=(600)=360000 and 2 AC. BC=2 x 600 x 700=840000
х BC=(700)=490000 Natural cosine C (56° 12') = .5563 850000
467292 382708 And /(382708)=618.63 yards=AB.
51. To find the distance between two inaccessible objects.
Let C and D be two inaccessible objects, the distance between which is required. Measure any
D base as AB; at A find the angles CAD, DAB; and at B find the angles ABC, CBD. In the triangle ABC, in which the angles and side AB are given, find CB; in like manner, in the triangle
B ABD, find BD. Then in the triangle CBD, of which the sides CB, BD, and the included angle BD are given; find the side CD.
Example. Let C and D be two houses on the further side of a river; required their distance from the following data, viz.; AB=1000 links, the CAD=42° 45', 4 DAB=54° 12', _ ABC=57° 33, and Z CBD=50° 19'.
1. In ABC, to find BC arith. com. log sin C (25° 30')
= 10.366016 log ein A (96° 57') 9.996797 log AB (1000)
3. log BC=2305.75 3.362813 2. In ABD, to find BD. arith, com. log sin D (17° 56')
10.511576 log sin A (54° 12') 9.909055 log AB (1000)
3. log BD-2634.09 3.420631 Now in the triangle CBD we know the two sides BC and BD, and the included angle CBD ; whence as in the last case we easily
; find CD=2120.95 links, the distance required.
52. Given the distances between three objects A, B, C, ana the angular distances between these objects at a station S in the same
horizontal plane ; to find the distance between that station and each of the objects.
Ist. Let the station S be without the triangle ABC, formed by lines connecting the three objects.
Construct the triangle ABC; at A make the angle DAB=the given angle CSB; at B, make the angle ABD=ASD. Then through the
C points A, B, D, describe a circle ; join DC, and produce it till it meet the circle in S, the station of the observer, and join SA, SB.
Then in the triangle ABC, of which the three sides are known, find the angle BAČ. In the triangle ABD, in which the angles and side AB are known, find the side AD. In the triangle CĂD, of which two sides AC, AD, and the included angle CAD are known, find the angle ACD. In the triangle ASC, of which the angles and the side AC are known, find the sides SA, SC. And in the triangle ASB, of which the angles and the side AB are known, find SB.
Example There are three objects A, B, C, (above figure) the distances of which from one another are as follows: AB=424, AC=213, BC=262 yards; I desire to know my distance from each of these objects, from a station S, without the triangle ABC, where I observed the angle ASC to be=13° 30', and BSC=29° 50' ?
Answer, BS=524.326, AS=605.712, CS=429.682 yards. 2. When the station S is within the triangle ABC.
Construct the triangle ABC; make the angles BAD, ABD, equal to the supplements of the given angles BSC,
D ASC, respectively. Through the points A, B, D, describe a circle,
C join DC, which will cut the circle in S, the station of the observer.
B This done, the calculation will be exactly as before
Example. The distances between three objects A, B, C, (above figure) are as follows: AB=267 yards ; AC=346, BC=209; and the angles at a point S within the triangle ABC, subtended by those distances, are 128° 40, 140°, 91° 20', respectively; what is the distance of the point S from each of these objects ?
Answer, AS=189.33, CS=178.85, BS=104.049 yards. 3. When the three objects are in one straight line.