N 40° 50' E to the right of the meridian, mark-
ing the point of angle measurement and joining
it with the point 6" by a straight line, which will
be the direction of the second course. The end
of this course is Sta. 10 + 89, and its length is
the difference between 1,089 and 520, which is
569 ft. This distance scale off from C, locat-
ing the point £ at Sta. 10 4- 89. In a similar
manner plat the remaining courses given in
the example. Write the bearing of each line
directly upon the line, taking care that the
bearings shall read in the same direction
as that in which the courses are being
run.
SURVEYING.
Ui
(643
(644
See Art. 1231 and Figs. 261 and 262.
For ^rst adjustment see Art. 1233. For
second adjustment see Art. 1234 and Fig. 263, and for
////W adjustment see Art. 1235 and Fig. 264.
(645) See Art. 1238 and Fig. 265.
(646) See Art. 1239 and Fig. 266.
(647) See Art. 1 240 and Fig. 267.
(648) See Art. 1 242 and Fig. 269.
(649) See Art. 1 242 and Fig. 269.
(650) To the bearing at the given line, viz. , N 55° 15' E,
we add the angle 15° IT', which is turned to the right. This
gives for the second line a bearing of N 70° 32' E.
(651) To the bearing of the given line, viz., N80°ll', we
add the angle 22° 13', which is the amount of change in the
direction of the line. The sum is 102° 24', and the direction
is 102° 24' to the right or cast of the north point of the com-
pass. At 90° to the right of north the direction is due east.
Consequently, the direction of the second line must be south
of east to an amount equal to the difference between
102° 24' and 90°, which is 12° 24'. Subtracting this angle
from 90°, the angle between the south and east points, we
have 77° 36', and the direction of the second line is S 77° 36' E.
The simplest method of determining the direction of the
second line is to subtract 102° 24' from 180° 00'. The differ-
ence is 77° 36', and the direction changing from N E to S E
gives for the second line a bearing of S 77° 36' E.
(652) To the bearing of the given line, viz. , N 13° 15' W,
we add the angle 40° 20', which is turned to the left. The
sum is 53° 35', which gives for the second line a bearing of
N 53° 35' W.
(653) The bearing of the first course, viz., S 10° 15' W,
is found in the column headed Mag. Bearing, opposite Sta. 0.
For the first course, the deduced or calculated bearing
must be the same as the magnetic bearing. At Sta. 4 -\- 40,
Un
SURVEYING.
an angle of 15° 10' is turned to the right. It is at once evi-
dent that if a person is traveling in the direction S 10° 15' W,
and changes his course to the right 15° 10', his course will
approach a due westerly direction by the amount of the
change, and the direction of his second course is found by
adding to the first course, viz., S 10° 15', the amount of such
Station.
Deflection.
Mag. Bearing.
Ded. Bearing.
54 + 25
49 + 20
L. 25° 14'
S 25° 40' W
S 25° 39' W
44 + 80
L. 10° 47'
S 50° 50' W
S 50° 53' W
33 + 77
R. 16° 55'
S61°45'W
S 6r 40' W
25 + 60
R. 24° 40'
S 44° 50' W
S 44° 45' W
16 + 20
L. 15° 35'
vS 20° 00' W
S 20° 05' W
8 + 90
R. 10° 15'
S 35° 50' W
S 35° 40' W
4 + 40
R. 15° 10'
S 25° 20' W
S 25° 25' W
S 10° 15' W
S 10° 15' W
change in direction. The sum is 25° 25', and the second
course S 25° 25' W. The needle at this point reads S 25° 20' W.
The difference between the magnetic bearing and the cal-
culated bearing may be owing to local attraction, but as we
cannot read the needle to within 10 minutes, we must gen-
erally ascribe small discrepancies to that cause. This cal-
culated bearing we write in its proper column opposite Sta.
4 + 40, where the change in direction occurred.
The next angle is 10° 15' to the right, which we add to the
previous calculated bearing S 25° 25' W, giving S 35° 40' W
for the calculated bearing of the third course, which extends
from Sta. 8 + 90 to 16 + 20. In a similar manner, the stu-
dent will calculate the remaining bearings, considering well
how the changes in direction will affect his relations to the
points of the compass. A plat of the notes to a scale of
400 ft. to the inch is given in Fig. 65.
SURVEYING.
US
64*^^^
^
Assume the starting point A,
Fig. Go, which is Sta. 0, and
through it draw a straight line.
The first angle, viz., 15° 10' to
the right, is turned at Sta. 4 + 40,
giving for the first course a length
of 440 ft. Scale off this distance
from A to a. scale of 400 ft. to
the inch, locating the point /?,
which is Sta. 4 -J- 40, and write on
the line its bearing S 10° 15' W,
as recorded in the notes. Pro-
duce A B to C, making B C greater
than the diameter of the protrac-
tor, and from B lay off to the
right oi B C the angle 15° 10'.
Join this point of angle measure-
ment with B hy 3. straight line,
giving the direction of the second
course. The end of this course is
at Sta. 8 -f 90, and its length is
the difference between 8 -|- 90 and
4 -i- 40, which is 450 ft. This dis-
tance we scale off from B, Sta.
4 -|- 40, locating the point Z>, and
write on the line its bearing of S
25° 30' W, as found in the notes.
We next produce B D to E, mak-
ing D E greater than the diameter
of the protractor, and at D lay
off the angle 10° 15' to the right
of D E, giving the direction of the
next course. In a similar man-
ner, plat the remaining notes given
in Example G53. The student in
his drawing will show the pro-
longation of only the first three
lines, drawing such prolongations
244
SURVEYING.
in dotted lines. In platting the remaining angles, he
will produce the lines in pencil only, erasing then, as
soon as the forward angle is laid oflf. Write the proper
station number in pencil at the end of each line as soon as
platted, and the angle with its direction, R. or L., before
laying off the following angle. Write the bearing of each
line distinctly, the letters reading in the same direction in
which the line is being run. The magnetic meridian is
platted as follows: The bearing of the course from Sta.
33 + 77 to Sta. 44 + 80 is S 61° 45' W, i. e., the course is
61° 45' to the left of a north and south line, which is the
direction we wish to indicate on the map. Accordingly, we
place a protractor with its center at Sta. 33 + 77 and its
zero on the following course, and read off the angle 61° 45'
to the right. Through this point of angle measurement and
Sta. 33 + 77 draw a straight line N S. This line is the
required meridian.
(654) Angle B = 39° 25'. From the principles of trigo-
nometry (see Art. 1 243), we have the following proportion:
sin 39° 25' : sin 60' 15' :: 415 ft. : side A B.
sin 60° 15' = .8682.
415 ft. X .8682 = 360.303 ft.
sin 39° 25' = .63496.
360.303 -^ .63496 = 567.442 ft., the side A B. Ans.
C
Fig. 66.
and the line A B produced as required.
(655) (See Fig. 66.) At
A we turn an angle B A (7 of
60° and set a plug at C 100'
from A. At C we turn an
angle A C B oi 60° and set a
plug at B 100' from C. The
point B will be in the line
A B, and setting up the in-
strument at B, we turn the
angle C B A = 60°. The
instrument is then reversed,
SURVEYING. 245
(656) See Art. 1245.
(657) See Art. 1246.
(658) See Art. 1246.
(659) See Art. 1248.
(660) See Art. 1 249.
(661) A 5° curve is one in which a central angle of
5° will subtend a chord of 100 ft. at its circumference. Its
radius is practically one-fifth of the radius of a 1° curve,
and equal to 5,730 ft. -^ o = 1,146 ft.
(662) The degree of curve is always twice as great as
the deflection angle.
(663) See Art. 1 249 and Fig. 282.
(664) Formula 90, C' = 2 i^ sin Z>. (See Art. 1250.)
(665) Formula 91, T= R tan ^ /. (See Art. 1251.)
(666) The intersection angle C E F^ being external to
the triangle A E C^ ys, equal to the sum of the opposite
interior angles A and C. A =22° 10' and C = 2S° 15'.
Their sum is 45° 25' = C E F. Ans.
The angle A E C^ISO" - (22° 10' + 23° 15') = 134° 35'.
From the principles of trigonometry (see Art. 1243),
we have
sin 134° 35' : sin 23° 15':: 253.4 ft. : sidey^^fi";
whence, side A E = 140.44 ft., nearly. Ans.
Also, sin 134° 35' : sin 22° 10' :: 253.4 ft. : side C E\
whence, side C E = 134.24 ft., nearly. Ans.
(667) We find the tangent distance T by applying
formula 91, r=i^ tan ^ /. (See Art. 1251.) From the
table of Radii and Deflections we find the radius of a 6° 15'
curve = 917.19 ft.; \ /=^^^^i^=17° 35'; tan 17° 35' =
.31G9. Substituting these values in the formula, we have
7^= 917.10 X .3169 = 290.66 ft. Ans.
(668) We find the tangent distance T by applying
formula 91, r= A' tan ^7. (See Art. 1251.) From the
246 SURVEYING.
table of Radii and Deflections we find the radius of a 3° 15'
14° 12'
curve is 1,763.18 ft.; ^ 1= = 7° OG'; tan 7° 06' =
.12456. Substituting these values in the above formula, we
have T= 1,763 X .12456 = 219.62 ft. Ans.
(669) See Art. 1252.
(670) The angle of intersection 30° 45', reduced to
decimal form, is 30.75°. The degree of curve 5° 15', reduced
to decimal form, is 5.25°. Dividing the intersection angle
30.75° by the degree of curve 5.25 (see Art. 1252), the
quotient is the required length of the curve in stations of
30 75°
100 ft. each. ^_^ = 5. 8571 full stations equal to 585. 71 ft.
(671) In order to determine the P. C. of the curve, we
must know the tangent distance which, subtracted from the
number of the station of the intersection point, will give us
the P. C. We find the tangent distance T by applying
formula 91, r= ^ tan I /. (See Art. 1251.) From the
table of Radii-and Deflections we find the radius of a 5° curve
is 1,146.28 ft. ; | /= ^IJ^ = 16° 33' ; tan 16° 33' = .29716.
P.C.5°R. Substituting these val-
A ^^ 't^340^3'^J^'^0-^37.8 ^ ues in formula 91, we
J6+97.17]lenSi^'^^O^J!33W have T = 1,146.28 X
' ^^%>\z> -29716 = 340.63 ft. In
I . ^3+59JZ^^' Fig. 67, let A B and
/ ^ C D \>Q. the tangents
/ which intersect in the
/ point E^ forming an
FIG- er. angle D E F=^^° 06'.
The line of survey is being run in the direction A B, and
the line is measured in regular order up to the intersection
point E, the station of which is 20 + 37.8. Subtracting the
tangent distance, B E= 340.63 ft. from Sta. 20+ 37.8, we
have 16 + 97.17, the station of the P. C. at B. The inter-
section angle 33° 06' in decimal form is 33.1°. Diviling
SURVEYING. 247
this angle by 5, the degree of the curve, we obtain the
33 1
length B G D oi the curve in full stations. -^ = 0.62
o
stations = 662 ft. The length of the curve, 663 ft., added
to the station of the P. C, viz., 16 + 97.17, gives 23 + 59.17,
the station of the P. T. at D.
(672) The given tangent distance, viz., 291.16 ft., was
obtained by applying formula 91, T—R tan ^ / (see
Art. 1251), /=20° 10', and | /= 10° 05', tan 10° 05' =
. 17783. Substituting these values in the above formula, we
291 16
have 291.16 = /? X .17783; whence, R = ^rp,= 1,637.29ft.
Ans.
The degree of curve corresponding to the radius 1,637.29
we determined by substituting the radius in formula 89,
50
R = -. — p: (see Art. 1249), and we have
sm D ^ "
1,637.29 = -^—n\ whence, sin D = ^-^^, = .03054.
sm D ' 1,637.29
The deflection angle corresponding to the sine .03054 is
1° 45', and is one-half the degree of the curve. The degree
of curve is, therefore, 1° 45' X 2 = 3° 30'. Ans.
(673) Formula 92, ^=^- (See Art. 1255 and
Fig. 283.)
(674) The ratio is 2; i. e., the chord deflection is double
the tangent deflection. (See Art. 1254 and Fig. 283.)
(675) As the degree of the curve is 7°, the deflection
angle is 3° 30' = 210' for a chord of 100 ft., and for a chord
210'
ot 1 ft. the deflection angle is ^^ = 2.1'; and for a chord of
48.2 ft. the deflection angle is 48.2x2.1' = 101.22'=l° 41.22'.
(676) The deflection angle for 100-ft. chord is ^-^ =
3° 07i' = 187.5', and the deflection angle for a 1-ft. chord is
248 SURVEYING.
^^^ = 1.875'. The deflection angle for a chord of 72.7 ft:
is, therefore, 1.875' X 72.7 = 136.31' = 2° 10.31'.
(677) We find the tangent deflection by applying
formula 93, tan def . = —75. (See Art. 1 255.) c = 50.
50^ = 2,500. The radius R of 5° 30' curve = 1,042.14 ft.
(See table of Radii and Deflections.) Substituting these
2 500
values in formula 93, we have tan def. = ^ ' , ^^ = 1.199
2,084.28
ft. Ans.
(678) The formula for chord deflections is d= -^. (See
Art. 1255, formula 92.) ^ = 35.2. 35.2' := 1,239.04. The
radius 7? of a 4° 15' curve is 1,348.45 ft. Substituting these
1 239 04
values in formula 92, we have d= ' ^,' , ^ = .919 ft. Ans.
1,348.45
(679) The formula for finding the radius ^ is 7? =
-.^,. (See Art. 1 249.) The degree of curve is 3° 10'. D,
sm V / o
3° 10'
the deflection angle, is -—— = 1° 35'; sin 1° 35' = .02763.
Substituting the value of sin D in the formula, we have R =
50
—-7^—-; whence, R = 1,809.63 ft. Ans.
The answer given with the question, viz., 1,809.57 ft.,
agrees with the radius given in the table of Radii and
Deflections, which was probably calculated with sine given to
eight places instead of five places, as in the above calculation,
which accounts for the discrepancy in results.
(680) In Fig. 68, let A B and A C represent the given
lines, and BC the amount of their divergence, viz., 18.22
B ft. The lines will form a
A ^.^ - -— '^ J^f^-^^ triangle ABC, of which
the angle A =. V. Draw
Fig. 68. a perpendicular from A to
D, the middle point of the base. The perpendicular will
SURVEYING. 249
bisect the angle A and form two right angles at the base of
the triangle. In the triangle A DB we have, from rule 5,
Art. 754, tan .5^/?=^. B A D = 30', BD=^^^^^ =
9.11 ft., and tan 30' = .00873. Substituting known values
in the
9.11 ft.
9 11
in the equation, we have .00873 = -4-^; whence, AD =
.00873 = l'»«-«^ ''■^"^-
By a practical method, we determine the length of the
lines by the following proportion:
.1.745 : 18.22 :: 100 ft. : the required length of line;
1 822
whence, length of line = ' = 1,044.13 ft. Ans.
The second result is an application of the principle of two
lines 100 ft. in length forming an angle of 1° with each
other, which will at their extremity diverge 1.745 ft.
(681) Degree of curve = |l^ = |i||l = 4°. Ans.
(682) See Arts. 1 264, 1 265, 1 266, and 1 267.
(683) See Arts. 1 269 -1 274.
(684) Denote the radius of the bubble tube by x\ the
distance of the rod from the instrument, viz., 300', hy d\
the difference of rod readings, .03 ft., by /r, and the move-
ment of the bubble, viz., .01 ft., by 5. By reference to Art.
1275 and Fig. 289, we will find that the above values have
the proportion h :S '.'. d \ x. Substituting known values
in the proportion, we have .03 : .01 :: 300 : x\ whence,
3
X = -r- — 100 ft., the required radius. Ans.
.03
(685) See Art. 1 277.
(686) See Art. 1 278.
(687) To the elevation 61.84 ft. of the given point, we
add 11.81 ft., the backsight. Their sum, 73.05 ft., is the
height of instrument. From this H. I., we subtract the fore-
250
SURVEYING.
sight to the T. P., viz., 0.49 ft., leaving a difference of 73. IG
ft., which is the elevation of the T. P. (See Art. 1279.)
(688) See Art. 1 280.
(689) See Art. 1281.
. (690) See Art. 1 282.
(691) See Art. 1286.
(692) See Art. 1289.
(693) See Art. 1 290.
(694) See Art. 1291.
(695) The distance between Sta. 66 and Sta. 93 is 2?
stations. As the rate of grade is + 1.25 ft. per station, the
total rise in the given distance is 1.25ft. X 27 = 33.75 ft.,
which we add to 126.5 ft., the grade at Sta. 66, giving 160.25
ft. for the grade at Sta. 93. (See Art. 1291.)
(696) See Art. 1292.
(697)
- 16.4'
56'
10.3'
73'
Contour 50.0 at 48.5 ft. to left of Center Line. £
Contour 40.0 at 94.0 ft. to left of Center Line, (j
Contour 30.0 at 128.0 ft. to left of Center Line.
+ 11.4' +8.8'
84'
96'
Contour 60.0 at 26 ft. to right of Center Line.
Contour 70.0 at 106.9 ft. to right of Center Line.
Elevation 76.7 at 180 ft. to right of Center Line
(698) The elevations of the accompanying level notes
are worked out as follows: The first elevation recorded in
the column of elevations is that of the denc/i mark, abbre-
viated to B. M. This elevation is 161.42 ft. The first rod
reading, 5.53 ft., is the backsight on this B. M., 2. plus read-
ing, and recorded in column of rod readings. This rod
reading we add to the elevation of the bench mark, to deter-
mine the height of instrument, as follows: 161.42 ft. + 5.53
ft. = 166.95 ft., the H. I. The next rod reading, which is
at Sta. 40, is 6.4 ft. The rod reading means that the sur-
face of the ground at Sta. 40 is 6.4 ft. below the horizontal
axis of the telescope. The elevation of that surface is,
therefore, the difference between 166.95 ft., the H. I., and 6.4
ft., the rod reading. 166.95 -6.4 = 160.55 ft. The— -ft. is a
SURVEYING.
251
fraction so small that in surface elevations it is the univer-
sal practice to ignore it, and the elevation of the ground at
Station.
Rod
Reading.
Height
Instrument.
Elevation.
Grade.
B. M.
+ 5.53
166.05
161.42
40
6.4
160.5
162.0
41
7.2
159.7
160.485
41 + 60
10.9
156.0
•
42
8.6
158.3
158.97
43
8.8
158.1
157.455
T. P. -
8.66
'
158.29
+
2.22
160.51
44
4.8
155.7
155.94
45
6.3
154.2
154. 425
46
8.8
151.7
152.91
47
9.9
150.6
151.395
48
11.1
140.4
149.88
T. P. -
11.24
140.27
+
3.30
152.57
49
4.7
147.9
148.365
50
7.1
145.5
146.85
51
8.7
143.9
145.335
52
9.8
142.8
143.82
53
10.9
141.7
142.305
T. P. -
11.62
140.95
Sta. 40 is taken at 160.5 ft. The rod reading at Sta. 41 is
7.2, which, subtracted from 166.05 ft., gives for that station
an elevation of 150.7. The remaining rod readings up to
252 SURVEYING.
and including that at Sta. 43, we subtract from the same
H. I., viz., 166.95. Here at a turning point (T. P.) of 8.66
ft. is taken and recorded in the column of rod readings.
This reading being a foresight is viinus, and is subtracted
from the preceding H. I. This gives us for the elevation of
theT. P., 166.95 ft. - 8.66 ft. = 158.29 ft., which we record in
the column of elevations. The instrument is then moved for-
wards and a backsight of 2.22 ft taken on the same T. P. and
recorded in the column. This is a plus reading, and is added
to the elevation of the T. P., giving us for the next H. T.
an elevation of 158.29 ft.+ 2.22 ft.= 160.51 ft. The next
rod reading, viz., 4.8, is at Sta. 44, and the elevation at that
station is the difference between the preceding H. I., 160.51,
and that rod reading, giving an elevation of 160.51 ft. — 4.8
ft. = 155.7 ft., which is recorded in the column of elevations
opposite Sta. 44. In a similar manner, the remaining ele-
vations are determined.
In checking level notes, only the turning points rod read-
ings are considered. It will be evident that starting from
a given bench mark, all the backsight or plus readings will
add to that elevation, and all the foresight or minus read-
ings will subtract from that elevation. If now we place in
one column the height of the B. M., together with all the
backsight or -|- readings, and in another column all the fore-
sight or — readings, and find the sum of each column, then,
by subtracting the sum of the — readings from the sum of
the -\- readings, we shall find the elevation of the last point
calculated, whether it be a turning point or a height of in-
strument. Applying this method to the foregoing notes
we have the following :
-f readings. — readings.
R M. 161.42ft. 8.66 ft.
5.5 3 ft. 11.2 4 ft.
2.2 2 ft. 11.6 2 ft.
^•^Q ^*- 31.52 ft.
17 2.4 7 ft.
31.52 ft.
X 4 0.9 5 ft.
SURVEYING.
253
The difference of the columns, viz., 140.95 ft., agrees
with the elevation of the T. P. following Sta. 53, which is
the last one determined. A check mark |/ is placed opposite
the elevation checked, to show that the figures have been
verified. The rate of grade is determined as follows: In one
mile there are 5,280 ft. = 52.8 stations. A descending grade
of 80 ft. per mile gives per station a descent of -r^-^" =1.515
52. o
ft. The elevation of the grade at Sta. 40 is fixed at 162.0
ft. As the grade descends from Sta. 40 at the rate of 1.515
ft. per station, the grade at Sta. 41 is found by subtracting
1.515 ft. from 162.0 ft., which gives 160.485 ft., and the
grade for each succeeding station is found by subtracting
the rate of grade from the grade of the immediately
preceding station.
A section of profile paper is given in Fig. 69 in which the
level notes are platted, and upon which the given grade line
^ FIO. 69. 60
is drawn. The profile is made to the following scales: viz.,
horizontal, 400 ft. = 1 in. ; vertical, 20 ft. = 1 in.
Every fifth horizontal line is heavier than the rest, and
each twenty-fifth horizontal line is of double xvcight. Every
tenth vertical line is of double iveight. The spaces between
the vertical lines represent JOO ft., and those between the
254 SURVEYING.
horizontal lines 1 ft. The figure represents 1,500 ft. in length
and 45 ft. in height. Assume the elevation of the sixth
heavy line from the bottom at 150 ft. The second vertical
line from the left of the figure is Sta. 40, which is written
in the margin at the bottom of the page. Under the next
heavy vertical line, ten spaces to the right, Sta. 50 is writ-
ten. The elevation of Sta. 40, as recorded in the notes, is
160.5 ft. We determine the corresponding elevation in
the profile as follows:
As the elevation of the sixth heavy line from the bottom
is assumed at 150 ft., 160.5 ft., which is 10.5 ft. higher, must
be 10^ spaces above this line. This additional space covers
two heavy lines and one-half the next space. This point is
marked in pencil. The elevation of Sta. 41, viz., 159.7 ft.,
we locate on the next vertical line and 9.7 spaces above the
150 ft. line. The next elevation, 156.0 ft., is at Sta. 41 -|-
60. This distance of 60 ft. from Sta. 41 we estimate by the
eye and plat the elevation in its proper place. In a similar
manner, we plat the remaining elevations and connect the
points of elevation by a continuous line drawn free-hand.
The grade at Sta. 40 is 162.0 ft. This elevation should be
marked in the profile by a point enclosed by a small circle.
At each station between Sta. 40 and Sta. 53 there has been
a descent of 1.515 ft., making a total descent between these
stations of 1.515 X 13 = 19.695. The grade at Sta. 53 will,
therefore, be 162.0 ft. — 19.695 ft. = 142.305 ft. Plat the
elevation in the profile at Sta. 53, and enclose the point in a
small circle. Join the grade point at Sta. 40 with that at
Sta. 53 by a straight line, which will be the grade 'line
required. Upon this line mark the grade — 1.515 per
100 ft.
-f 11°
(699) ^ ^
120'
Nine 5-foot contours are included within the given slopes,
as follows:
Contour 70.0 at 31 5 ft. to left of Center Line. «;
Contour 65 at 63.0 ft, to left of Center Line. '2
Contour 60.0 at iM..5 ft. to left of Center Line, fc
Contour M.O at 133 ft to left of Center Line. =
Elevation 50.7 at 182.0 ft. to left of Center Line."
Contour 80.0 at 25.5 ft. to right of Center Line.
Contour 85.0 at 51 .0 ft. to right of Center Line.
Contour 90.0 at 70.5 ft to right of Center Line,
("ontour 95.0 at 102.0 ft. to right of Center Line.
Elevation 98.5 at 1200 ft. to right of Center Line.
SURVEYING. 255
(TOO) 1.745 ft. X 3 = 5.235 ft., the vertical rise of a 3°
slope in 100 ft., or 1 station and -——- =1.91 stations = 191 ft.
5.2o5
(701) (See Question 701, Fig. 15.) From the instru-
ment to the center of the spire is 100 ft. + 15 = 115 ft., and
we have a right triangle whose base A D = 115 ft. and angle
A is 45° 20'. From rule 5, Art. 754, we have tan 45° 20' =
side/?/? , , ^,,,v^ side BD „ „ ,,„.►.
— -— — ; whence, 1.01170 = -— ; ox B D — 11(5.345
115 llo
feet. The instrument is 5 feet above the level of the base;
hence, 116.345 ft. + 5 ft. = 121.345, the height of the spire.
(702) Apply formula 96.