I.

PACING
1. A 45m course, AB, on level ground was paced by a surveyor for
the purpose of determining his pace factor. The number of paces
for each trial taken are shown in the accompanying tabulation.

Requirements:
a. Determine his pace factor
b. If the surveyor then took 771,770,768,770,772 and 769 paces in
walking an unknown distance CD, what is the length of the line?
c. Determining relative precision.

TRIAL LINE TAPED MEAN

DISTANCE
1 AB 50
2 BA 53
3 AB 51
45 52
4 BA 53
5 AB 52
6 BA 53

SOLUTION:
a. Given: L = 45 m; n1 = 6

SUM1 = (50 + 53 + 51 + 53 + 52 + 53) = 312 paces

M1 = SUM1 / n1 = 312 / 6 = 52 paces
PF = L / M1 = 45 / 52 = 0.865 m / paces
b. n2 = 6
SUM2 = (771 + 770 + 768 + 770 + 772 + 769) = 4620 paces
M2 = SUM2 / n2 = 4620 / 6 = 770 paces
PD = L / M2 = 770 (0.865 / pace) = 666.1 m

c. TD = 667m
PD = 666.1 m
RP = (TD - TP) / TD = (667 – 666.1) / 667
= 0.9 / 667
 = 1/741 say 1/700
2. In five trials of walking along a 90m course on fairly level
ground, a pacer for a survey party counted 51, 52.5 , 51.5 , 52.5 ,
and 51.5 strides respectively. He then started walking an unknown
distance XY in four trials which were recorded as follows: 88.5,
89, 88, and 87 strides. Determine the following:
a. Pace factor of the pacer
b. Length of line XY
c. Percentage of error in the measurement if the taped length of XY is
150.5 meters

a. Given: L = 90m

n1 = 5
SUM1 = (51+52.5+51.5+52.5+51.5) = 259 strides or 518 paces
M1 = SUM1 / n1 = 518 / 5 = 103.6 paces
PF = L / M1 = 90 / 103.6 = 0.869 m/ paces

b. n2 = 4
SUM2 = (88.5+89+88+87) = 352.5 strides or 705 paces
M2 = SUM2 / n2 = 705 / 4 = 176.25 paces
PD = L / M2 = 770 (0.865 / pace) = 666.1 m

c. TD = 150.5m
PD = 153.2m

Percentage Error = (TD-PD / TD)(100%)

= (150.5 – 153.2 / 150.5)(100%)
= 1.79%
II. DOUBLE AREA METHOD
1) In figure shown below and the accompanying tabulation are given the
adjusted latitudes and adjusted departures of a closed traverse.
Calculate the area by the DMD and DPD method.

AREA BY DOUBLE MERIDIAN DISTANCE

Adjusted latitude Adjusted Departure

LINE
(+N) (-S) (+E) (-W)
AB 490.71 47.27
BC 587.12 608.89
CD 327.41 786.78
DE 1002.76 218.32
EF 122.67 1116.62
FA 375.01 544.64
SUM 1452.64 -1452.84 1661.26 -1661.26
a. Computation of DMD (Refer to the rules of the computing DMD)

DMDab = 47.27

DMDbc = 47.27 + 47.27 + 608.89 = 703.43

DMDcd = 703.43 + 608.89 + 786.78 = 2099.10

DMDde = 2099.10 + 786.78 + 218.32 = 3104.20

DMDef = 3104 + 218.32 -1116.62 = 2205.90

DMDfa = 2205.90 – 1116.62 -544.64 = 544.64

b. Computation of double areas: DOUBLE AREA = DMD X Adj Lat.

DAab = 47.27 (490.71) = + 23,195.86

DAbc = 703.43 (587.12) = + 412,997.82

DAcd = 2099.10 (-327.41) = - 687,266.33

DAde = 4104.20 (-1002.76) = - 3,112,767.59

DAef = 2205.90 (-122.67) = - 270,597.75

DAfa = 544.64 (375.01) = + 204,245.45

∑DA = - 3,430,192.54

THEREFORE:

2 X AREA = - 3430192.54

AREA = - 1,715,096.27 sq. m (negative sign is disregarded)

AREA BY DOUBLE PARALLEL DISTANCE. Using the given data in
problem no. 1, determine the area of the closed traverse by DPD method.

a. Computation of DPD’s (Refer to rules for computing DPD)

DPD ab = 490.71

DPDbc = 490.71 + 490.71 + 587.12 = 1,568.54

DPDcd = 15668.54 + 587.12 - 327.41 = 1828.25

DPDde = 1828.25 – 327.41 -1002.76 = 498.08

DPDef = 498.08 -1002.72-122.67 = - 627.35

DPDfa = -627.35 – 122.67 + 375.01 = - 375.01

b. Computation of Double Areas (DOUBLE AREA = DPD X ADJ.

LATITUDE)

DA ab = 490.71 x 47.27 = + 2,395.86

DA bc = 1568.54 x 608.89 = + 955,068.32

DA cd = 1828.25 x 786.78 = + 1,438,430.54

DA de = 498.08 x 218.32 = + 108,740.83

DA ef = 627.35 x (-1116.62) = + 700,511.56

DA fa = 375.01 x (-544.64) = + 204,245.45

∑DA = 3,430,192.56

THEREFORE:

2 X AREA = 3,430,192.56

AREA = 1,715,096.28 SQ.M.

 2) Figure shown the example bearing traverse with the given computed
adjusted latitudes and departures.

Adjusted
Line Lat (ft) Dep (ft)
AB -176.386 -438.574
BC +203.382 -73.105
CD +192.340 +198.635
DE -219.336 +313.044

Starting with line AB and use Equation (G-4) to compute the DMDs.

Multiply DMDs by Lats; add the products

Adjusted
Line Lat (ft) Dep (ft) DMD (ft) DMD x Lat (sq ft)
AB -176.386 -438.574 -438.574 +77,358.3
BC +203.382 -73.105 -950.253 -193,264.4
CD +192.340 +198.635 -824.723 -158,627.2
DE -219.336 +313.044 -313.044 +68,661.8
sum: -205,871.5
Compute the area:
III. STADIA METHOD
1) An automatic level with an internal focusing telescope was set up
somewhere at mid length of a long span steel bridge. The rod
readings tabulated below were observed on a stadia rod held
successively at the vicinity of the concrete abutments in the
southern and northern approaches of the bridge. If the stadia
interval factor of the instrument is 98.5, determine the length of the
bridge.
HAIR READINGS
ROD POSITION
UPPER (a) MIDDLE (b) LOWER (b)
ROD AT
SOUTHERN 2.98m 1.68m 0.38m
APPROACH

ROD AT
NORTHERN 3.54m 2.02m 0.49m
APPROACH

SS = 2.98 - 0.38 = 2.60

Dn = 98.5 (3.05) = 300.4m
DS = 98.5 (2.60) = 256.1m
D = 256.1 + 300.4 = 556.5m
Sn = 3.54 - 0.49 = 3.05m

2) Compute the elevation of point A

Solution:
ElevA = HI + Vertical Elevation
Difference – Rod Reading
ElevA = HI + {s * (sin 2α) / 2} –
Rod Reading
ElevA = 105.00 + {100 * (sin 2(-
12)) / 2} – 8.00
ElevA = 105.00 + (-20.34) – 8.00
= 76.66’
IV. COMPASS AND TRANSIT RULE
1.) COMPASS RULE. Given in the accompanying tabulation are the
observe data for a traverse. Compute linear closure and precision and
adjust the latitudes and departures using the Compass Rule. Tabulate
values accordingly.

COURSE DISTANCE (m) BEARING

AB 379.25 S 36°00'00" E
BC 408.90 S 75°23'18" W
CD 298.65 N 14°43'29" W
DA 276.55 N 63°59'20" E
Solution:
A
S36°E

N63o59'20"W

B
S75°W

N14o43'29"W

a.) Computing latitudes: Lat = Length x cos α

LatAB = 379.25 cos 36°00' = - 306.82 m
LatBC = 408.90 cos 75°23'18” = - 103.15 m
LatCD = 298.65 cos 14°43'29" = + 288.84 m
LatDE = 276.55 cos 63°59'20" = + 121.28 m
b.) Computing departures: Dep = Length x sin α
DepAB = 379.25 sin 36°00' = + 222.92 m
DepBC = 408.90 sin 75o23’18” = - 395.68 m
DepCD = 298.65 sin 14o43’29”= - 75.923 m
DepDE = 276.55 sin 63o59’20” = + 248.54 m
 c.) Tabulated Solution:

COMPUTED COMPUTED
COURSE DISTANCE BEARING LATITUDE DEPARTURE
+N -S +E -W
S 36°00'00"
AB 379.25 - 306.82 222.918 -
E
S 75°23'18"
BC 408.90 - 103.152 - 395.675
W
N
CD 298.65 14°43'29" 288.842 - - 75.909
W
N
DA 276.55 121.28 - 248.538 -
63°59'20" E
1,363.35 m + 410.122 - 409.972 +471.456 -471.584
SUMS (Length of
(Σ NL) (Σ SL) (Σ ED) (Σ WD)
Traverse)

d.) Determining total closure in latitude and departure.

CL = ΣNL + ΣSL
= +410.122 + (-409.972)
= +0.150 m (Total Closure in Latitude)
CD = ΣED + ΣWD
= +471.456 + (-471.584)
= +0.128 m (Total Closure in Departure)
e.) Compute linear closure and precision
LEC = √(𝐶𝐷 )2 + (𝐶𝐿 )2
LEC = √(0.15)2 + (−0.128)2
LEC = 0.197
0.197 1
PREC = =
1363.35 6,920

f.) Adjusting the Latitudes:

𝐶
Adj. Lat = Lat + L ( 𝐿 )
𝐿
−(+0.150)
Adj. Latab = - 306.820 + 379.25 ( ) = -306.862 m
1363.35
−(+0.150)
Adj. Latbc = - 103.152 + 408.90 ( ) = + - 103.197 m
1363.35
 −(+0.150)
Adj. Latcd = + 288.842 + 298.65 ( ) = +288.809 m
1363.35
−(+0.150)
Adj. Latda = + 121.820 + 276.55 ( ) = +121.250 m
1363.35
g.) Adjusting the Departures:
𝐶
Adj. Dep = Dep + L ( 𝐷 )
𝐿
−(−0.128)
Adj. Depab = + 222.918 + 379.25 ( ) = +222.954 m
1363.35
−(−0.128)
Adj. Depbc = - 395.675 + 408.90 ( ) = -395.637 m
1363.35
−(−0.128)
Adj. Depcd = - 75.909 + 298.65 ( ) = -75.881m
1363.35
−(−0.128)
Adj. Depda = + 248.538 + 276.55 ( ) = +248.564 m
1363.35

h.) Tabulated Solution:

CORRECTION ADJ. LATITUDE ADJ. DEPARTURE

COURSE
Latitude Departure +N -S +E -W
AB 1.67 1.79 306.862 222.954
BC 2.86 3.08 103.197 395.637
CD 2.88 3.09 288.809 75.881
DA 3.43 3.69 121.250 248.564
SUM 16.81 18.08 +410.059 -410.059 +471.518 -471.518

2.) TRANSIT RULE. Given in the accompanying tabulation are the

observe data for a traverse obtained from a transit – tape survey.
Determine the latitudes and departures of each course and balance
these quantities by employing the transit rule. Tabulate values
accordingly.
CD
855.45 m
BC
850.62 m

AB DE
495.85 m 1,020.87 m

FA
660.08 m
EF
1,117.26 m
 AZIMUTH AZIMUTH
COURSE LENGTH FROM COURSE LENGTH FROM
NORTH NORTH
AB 495.85 m 5o30’ DE 1,020.87 m 167o35’
BC 850.62 m 46o02’ EF 1,117.26 m 263o44’
CD 855.45 m 112o22’ FA 660.08 m 304o51’

Solution:
a.) Computing latitudes:
Latab = 495.85 cos 5o30’ = +493.57 m
Latbc = 850.62 cos 46o02’ = +590.53 m
Latcd = 855.45 cos 112o22’ = -325.53 m
Latde = 1,020.87 cos 167o35’ = -996.99 m
Latef = 1,117.26 cos 263o44’ = -121.96 m
Latfa = 660.08 cos 304o51’ = +377.19 m
b.) Computing departures:
Depab = 495.85 sin 5o30’ = +47.53 m
Depbc = 850.62 sin 46o02’ = +612.23 m
Depcd = 855.45 sin 112o22’ = +791.09 m
Depde = 1,020.87 sin 167o35’ = +219.51 m
Depef = 1,117.26 sin 263o44’ = -1,110.58 m
Depfa = 660.08 sin 304o51’ = -541.70 m
c.) Tabulated Solution:
AZIMUTH COMPUTED COMPUTED
COURSE DISTANCE FROM LATITUDE DEPARTURE
NORTH +N -S +E -W
AB 495.85 m 5o30’ 493.57 47.53
BC 850.62 m 46o02’ 590.53 612.23
CD 855.45 m 112o22’ 325.53 791.09
DE 1,020.87 m 167o35’ 996.99 219.51
EF 1,117.26 m 263o44’ 121.96 1,110.58
FA 660.08 m 304o51’ 377.19 541.70
+1,670.3 -
SUMS 5,000.13 m +1,461.29 -1,444.48
6 1,652.28
d.) Determining total closure in latitude and departure.
CL = ΣNL + ΣSL
= +1,461.29 + (-1,444.48)
= +16.81 m (Total Closure in Latitude)
CD = ΣED + ΣWD
= +1,670.36 + (-1.652.28)
= +18.08 m (Total Closure in Departure)
𝐶𝐿
e.) Determining corrections for latitude: Kl = ; cl = Lat (Kl)
ΣNL− ΣSL
16.81 16.81
Kl = = = 0.000579
+1,461.29−(−1,444.48) 2,905.77
Cab = 493.57 (0.000579) = 2.86 m
Cbc = 590.53 (0.000579) = 3.42 m
Ccd = 325.53 (0.000579) =1.88 m
Cde = 996.99 (0.000579) = 5.77 m
Cef = 121.96 (0.000579) = 0.70 m
Cfa = 377.19 (0.000579) = 2.18 m
𝐶𝐷
f.) Determining corrections for departure:K2 = ;cd = Dep(Kl)
ΣED− ΣWD
18.08 18.08
K2 = = = 0.00544
+1,670.36−(−1,652.28) 3,322.64
Cab = 47.53 (0.00544) = 0.26 m
Cbc = 612.23 (0.00544) = 3.33 m
Ccd = 791.09 (0.00544) = 4.31 m
Cde = 219.51 (0.00544) = 1.19 m
Cef = 1,110.58 (0.00544) = 6.04 m
Cfa = 541.70 (0.00544) = 2.95 m
g.) Determining adjusted latitudes:
Adj. Latab = + (493.57 - 2.86) = +490.71 m
Adj. Latbc = + (590.53 – 3.42) = +587.11 m
Adj. Latcd = - (325.53 + 1.88) = -327.41 m
Adj. Latde = - (996.99 + 5.77) = -1,002.76 m
Adj. Latef = - (121.96 + 0.70) = -122.66 m
Adj.Latfa = + (377.19 – 2.18) = +375.01 m
h.) Determining Adjusted departures:
Adj. Depab = + (47.53 – 0.26) = + 47.27 m
Adj. Depbc = + (612.23 – 3.33) = + 608.90 m
Adj. Depcd = + (791.09 – 4.31) = +786.78 m
Adj. Depde = + (219.51 – 1.19) = +218.32 m
 Adj. Depef = - (1,110.58 + 6.04) = -1,116.62 m
Adj. Depfa = - (541.70 + 2.95) = -544.65 m
i.) Tabulated Solution:

ADJUSTED
CORRECTION ADJUSTED LATITUDE
COURSE DEPARTURE
LAT DEP +N -S +E -W
AB 2.86 0.26 490.71 47.27
BC 3.42 3.33 587.11 608.90
CD 1.88 4.31 327.41 786.78
DE 5.77 1.19 1,002.76 218.32
EF 0.70 6.04 122.66 1,116.62
FA 2.18 2.95 375.01 544.65

SUMS 16.81 18.08 +1,452.83 -1,452.83 +1,661.27 -1,661.27

V. STADIA LEVELING
1.) Complete the stadia level notes shown Complete the stadia level notes
shown below and perform arithmetic check. Assume that K Assume
that K 100 = and C 0= 0.

BACKSIGHT FORESIGHT
DE ELEV
STA RR VD RR VD
S(m) θ S(m) θ (m) (m)
(m) (m) (m) (m)
BM1 1.245 -4°25’ 2.42 75.0
TP1 2.044 3°20’ 1.08 1.515 8°18’ 1.55
BM2 1.438 -3°25’ 3.06

Solution:
BACKSIGHT FORESIGHT
DE ELEV
STA RR VD RR VD
S(m) θ S(m) θ (m) (m)
(m) (m) (m) (m)
BM1 1.245 -4°25’ 2.42 -9.56 75.0
TP1 2.044 3°20’ 1.08 11.86 1.515 8°18’ 1.55 21.64 32.07 107.07
BM2 1.438 -3°25’ 3.06 -8.55 -22.39 84.68

DE = RRBS −VDBS +VDFS − RRFS

DE = 2.42 – (-9 56) + 21.64 – 1.55
DE = 32.07 m
DE = 1.08 - 11.86 + (-8.55) – 3.06
DE = -22.39 m

2.) A line of third order levelling is run by theodolite, using tacheometric

methods with a staff held vertically. The usual three staff readings, of
center and both stadia hairs, are recorded together with the vertical
angle (V.A.) A second value of height difference is found by altering
the telescope elevation and recording the new readings by the vertical
circle and center hair only. The two values of the height differences
are then mean. Compute the difference in height between the points A
and B from the following data:

The stadia constants are multiplying constant = 100, additive constant

=0
BACKSIGHTS FORESIGHTS REMARKS
(all
V.A. STAFF V.A. STAFF measurements
in ft)
+0o02’ 6.20
4.65
Point A
3.10
+0o20’ 6.26
-0o18’ 10.20
6.60
Point B
3.00
0o00’ 10.37

Height difference between the two ends of the theodolite ray = 100s
sin0 cos 0, where s = stadia intercept and 6 = V.A.)
V = 100 s sin 6 cos 6
= 50 s sin 2 6
To A, V = 50(6-20 - 3*10) sin 0°04'
= +0-18 ft
Difference in level from instrument axis +0.18
-4.65
-4.47
Check reading
V = 50(3-10) sin0°40'
= +1.80
Difference in level from instrument axis + 1.80
-6.26
-4.46
mean -4.465
To B,
V = 50(10-20 -3-00) sin 0°36'
= -3-76
Difference in level from instrument axis - 3.76
- 6.60
-10.36
Check level -10.37
mean -10.365
Difference in level A - B - 10.365
- 4.465
- 5.900 ft
 VI. AREA WITH IRREGULAR BOUNDARIES (BY
TRAPEZOIDAL RULE OR SIMPSON’S RULE)
1.) A series of perpendicular offsets were taken from a transit line to an
irregular boundary. The following offsets were taken at 15 m intervals
from a survey line to an irregular boundary line: 3.50, 4.30, 6.75, 5.25,
7.50, 8.80, 7.90, 6.40, 4.40, and 3.25 m. By the trapezoidal rule and
simpson’s one third rule, find the area included between the transit
line, the curve boundary and the end offsets.
Determining area of an irregular boundary.

8.80 m
9.30 m
hh6
6 7.90 m
7.50 m
6.75 m h5 h7 6.40 m
h3 5.25 m h8
4.30 m h4 4.40 m
3.50 m h2 h9 3.25 m
h1 h10

end offset end offset

d d d d d d d d d

Solution:
a.) By Trapezoidal Rule:
ℎ1 + ℎ1𝑜
AREA = d ( + h2 + h 3 + h4 + h 5 + h 6 + h 7 + h 8 + h 9 )
2
3.50 + 3.25
AREA = 15 ( + 4.30 + 6.75 + 5.25 + 7.50 + 8.80 + 7.90+
2
6.40 + 4.40)
AREA = 820.125 m2
b.) By Simpson’s One Third Rule:
𝑑 𝑑
AREA = [(h1 + h9) + 2(h3 + h5 + h7) + 4(h2 + h4 + h6 + h8)] +
3 2
(h9 + h10)
15
AREA = [(3.50 + 4.40) + 2(6.75 + 7.50 + 7.90) + 4(4.30 + 5.25
3
15
+ 8.80 + 6.40)] + (4.40 + 3.25)
2
AREA = 813.375 m2

2.) From a transit line to the edge of a river, a series of perpendicular

offsets are taken. These offsets are spaced 4.0 meters apart and were
measured in the following order: 1.5 , 2.4 , 3.5 , 6.6 , 9.5 , 8.4 , 4.8 ,
6.1 and 3.3 meters. By Simpson’s One – third rule and trapezoidal
rule, compute the area included between the transit line, the river’s
edge and the line offsets.
 Determining area of a curved boundary.
9.50 m
8.5
h5
h5 8.40 m
7.4 h6
h6
6.60 m
h4 5.6 6.10 m
h4
4.80 m h5.1
8
h
8
h7
3.8
h7
3.50 m 3.30 m
h3 2.5 2.3 h9
2.40 m h3 h9
1.50 m h1.4
2
h2
h1
0.5
h1

end offset end offset

d d d d d d d d

Solution:
a.) Simpson’s One Third Rule:
𝑑
AREA = [h1 + h9 + 2(h3 + h5 + h7) + 4(h2 + h4 + h6 + h8)]
3
4
AREA = [1.50 + 3.30 + 2(3.50 + 9.50 + 4.80) + 4(2.40 + 6.60 +
3
8.40 + 6.10)]
AREA = 179.20 m2
b.) Trapezoidal Rule:
ℎ1 +ℎ9
AREA = d [( ) + h2 + h3 + h4 + h5 + h6 + h7 + h8]
2
1.50 +3.30
AREA = 4 [( ) + 2.40 + 3.50 + 6.60 + 9.50 + 8.40 + 4.80 +
2
6.10]
AREA = 174.80 m2
 VII. MISSING SIDES AND/OR BEARING OF CLOSED
TRAVERSE
1.) Given the following descriptions of a four - sided lot.

LINE BEARING DISTANCE

AB N 30” 30’ E 56.5m
BC N 75” 30’ W 46.5m
CD S 45” 30’ W 87.5m
DA - -

a.) What is the length of line DA?

b.) What is the bearing of line DA?
c.) Compute the area of the enclosed traverse.
SOLUTION:
a.) Length of DA:
LINE LAT DEP
AB +48.68 +28.68
BC +11.64 -49.05
CD -61.33 -62.41
DA +1.01 +78.82

Distance DA = √(+1.01)² + (+78.82)²

Distance DA = 78.83 m
b.) Bearing of DA:
Tan Ɵ = 78.82
1.01
Ɵ = 78” 02’
Bearing of DA = N 78” 02’ E
c.) Area of enclosed traverse:
LINE LAT DEP DMD PDA
AB +48.68 +28.68 +28.68 +1396.14
BC +11.64 -45.09 +12.27 142.82
CD -61.33 -62.41 -95.23 +5840.45
DA +1.01 +78.82 2A = 7299.81

2A = 7299.81
A = 3649.91 m²

2.) A closed traversed shows tabulated values of latitudes and

departures.

LINES LATITUDE DEPARTURE

1-2 +84.60 -
2-3 +95.32 -56.11
3-4 +62.66 -57.52
4-5 -48.16 -31.40
5-6 -43.04 +59.70
6-1 - +47.63

4 a.) Compute the DMD of line 3-4

b.) Compute the length of line 6-1
5 3 c.) Compute the bearing of line 6-1
6
2

1
SOLUTION:
a.) DMD of line 3-4:

Latitude of (6-1):
Σ Lat = 0
+ 84.60 + 95.32 + 62.66 - 48.16 - 43.04 – y = 0
y = - 151.38

Departure of (1-2):
Σ(Dep) = 0
x - 56.11 - 57.52 - 31.40 + 59.70 + 47.63 = 0
x = 37.70

LINES LATITUDE DEPARTURE DMD

1-2 +84.60 +37.70 37.70
2-3 +95.32 -56.11 19.29
3-4 +62.66 -57.52 -94.34
4-5 -48.16 -31.40 -183.26
5-6 -43.04 +59.70 -154.96
6-1 -151.38 +47.63 -47.63

DMD of line (3-4) = -94.34

b.) Length of line 6-1
(D)² = (Lat)² + (Dep)²
(D₆₋₁)² = (-151.38)² + (+47.63)²
D₆₋₁ = 158.70 m
c.) Bearing of line 6-1
Tan Ɵ = 43.63
151.38
Ɵ = 17” 29’
Bearing is S 17” 29’ E
 VIII. DIVIDING OF AREA OF CLOSED TRAVERSE
1.) A triangular lot has the following Azimuths and distances.
LINES AZIMUTH DISTANCE
1-2 180” 00’ ---
2-3 300” 00’ ---
3-1 40” 00’ 960.22

a.) The lot is to be divided such that the area of the southern portion
would be 210,000m². Compute the position of the other end of the
dividing line, if the line starts at corner 3 of the lot. Express the
distance from corner 1.
b.) What is the length of the dividing line?
c.) Compute the azimuth of the dividing line.
SOLUTION:
a.) Location of x from corner 1
𝑥(960.22)(𝑠𝑖𝑛40°)
210,000 =
2

x = 680.47

b.) Length of dividing line

(y)² = (680.7)² + (960.22)² - 2(680.47)(960.22) cos (40°)
y = 619.67m
c.) Bearing of dividing line:

680.47 619.67
=
sin 𝜃 sin 40 °
Ɵ = 44” 54’
Bearing = S 44” 54 W
Azimuth = 84” 54’
 2.) The following data of a given 4 sided lot.
LINES BEARING DISTANCES (m)
AB N 20” E ?
BC Due East 200
CD S 30” W ?
DA Due West 300
a.) Find the area of the lot in m².
b.) Find the length of the dividing line (EF) that is parallel to line DA
which will divide the lot into two equal areas.
c.) Determine the location of one end of the dividing line E from corner
A along line AB.
SOLUTION
a.) Area of lot:

(300)2 −(200)2
A=
2 ((cot 70 )+cot 58)

A = 25,282.16 m2
c.) Length of dividing line:
√(1)(300)2 +(1)(200)2
x=
1+2
x = 254.95 m
c.) Distance AE:
a = 300 – 254.95
a = 45.05
𝐴𝐸 45.05 AE = 48.48 m
=
sin 58 ° sin 52°