SURVEY COMPUTATION 2

JABATAN KEJURUTERAAN AWAM

CG203-SURVEY COMPUTATION 2

NOOR FAIZAH BINTI ZOHARDIN

Table of Contents

1.0 MISSING LINE IN CLOSED TRAVERSE
Calculation of closed traverse for one or two lines without data such as bearing, data
or both.

2.0 BOUNDARY PROBLEMS
Solving boundary problems related to rectangular and straight boundaries.

3.0 AREA DIVISION
Problem solving related to area and lots division.

4.0 ROUTE PROBLEMS
Problem solving related to road junction, route with different widths and new
routes via old lots.

5.0 THREE POINT AND THREE DISTANCES PROBLEMS
Types of coordinates and calculation of surveyed area using coordinate method

PRAKATA

Alhamdullah, segala puji bagi Allah, tuhan pencipta sekalian alam.

Buku ini dihasilkan bagi memberi sedikit panduaan kepada pelajar-pelajar Diploma Ukur
Tanah Politeknik yang mengambil subjek CG203 Survey Computation 2.

Ia dihasilkan dengan ilustrasi dan jalan kerja yang mudah difahami untuk menarik minat
pelajar mempelajari Ilmu pengiraan dalam bidang Ukur dengan lebih mendalam. Ia juga
dihasilkan berdasarkan Kurikulum terbaharu Politeknik Malaysia, Kementerian Pengajiaan
Tinggi

Adalah diharap buku ini dapat menjadi panduaan bagi pelajar untuk memahami serta
mencintai ilmu ukur.

NOOR FAIZAH BINTI ZOHARDIN
PegawaiPendidikanPengajiaanTinggi
JabatanKejuruteraanAwam
Politeknik Kuching Sarawak
CETAKKAN PERTAMA :JUN 2012
SEMAKAN PERTAMA : DISEMBER 2012
SEMAKAN KEDUA : JUN 2013
CETAKKAN KEEMAPAT : DISEMBER 2013

FORMULA

1. Latitude = distance x cos θ
2. Departure = distance x sin θ
3. Distance = (latit)2 + (depart)2
4. Angle = tan−1 depart

latit

Bearing = value of θ Bearing = 180° - θ

Bearing = 180° + θ Bearing = 360° - θ

Triangulation 1. c² = a² + b² - 2ab cos C
Road secant 2. cos = 2+ 2− 2

2

3. area = 1 absin C

2

4. area = s(s − a)(s − b)(s − c)
s= a + b + c

2
5. sin A = sin B

ab

α = bearing BA - bearing BC
2

θ = bearing AD – bearing AE
cot α = L1 cosec − cot

L2

Tree distance and tree point problems

BC(AB + BC + CD) = sin b(sin a + sin b + sin c)
( AB)( BC ) (sin a)(sin c)

Tan φ = a sin q
b sin p

tan x − y = tan( − 45) tan (x + y)
22

CHAPTER 1:
MISSING LINE IN
CLOSE TRAVERSE

CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
SOLVE MISSING DATA FOR ADJUSSENT LINE

EXAMPLE 1
I) Find bearing for line CD and line DE

STEP BY STEP
1) First ,joint point from C to E, after that, calculate bearing and distance for that line.

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
Calculate bearing and distance CE DEPARTMENT OF CIVIL ENGINEERING

LINE BEARING DISTANCE LATIT DEPART Latit = distance x cos θ
EF 281⁰36’23” 61.448 12.363 -60.192 Depart = distance x sin θ
FG 353⁰39’11” 53.430 53.103 -5.907
GA 50⁰05’59” 49.496 31.749 37.971 Calculate total of latit
AB 94⁰17’04” 45.129 -3.371 45.003 and depart
BC 127⁰33’36” 49.160 -29.968 38.970
CE -63.876 -55.845
check 0.000 0.000

Distance CE = √63.8762 + 55.8452
= 84.846


tan =

= tan−1 55.845
63.876

= 41°09′44"

Latit (N/S) Depart (E/W)
-63.876 -55.845

Bearing CE = 180⁰ + 41°09′44" Identify the bearing
= 221⁰ 09’ 44” position base on latit and
depart data.

LINE BEARING DISTANCE LATIT DEPART
CE 221⁰ 09’ 44” 84.846 -63.876 -55.845

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

60.793

Calculate bearing of CD Angle CDE is obtuse
angle , so the real angle
i. calculate angle of c using cos method must calculate like that

2 = 2 + 2 − 2( )( ) cos D = 180⁰ -83°12′28"

60.7932 = 84.8462 + 52.4322 − 2(84.846)(52.432) cos =96⁰ 47’ 32”
C = 45⁰ 21’ 20”

ii. bearing CD = 221⁰ 09’ 44” - 45⁰ 21’ 20”
= 175⁰ 48’ 24”

iii. Calculate bearing DE using sin method 60.793
sin 45°21′20" sin
60.793 = 84.846
= 83°12′28"

So, bearing DE = (180⁰ + 175⁰ 48’ 24”) - 96⁰ 47’ 32”

= 259⁰ 00’ 52”
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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
EXAMPLE 2
I) Find bearing for line AG and line AB

Calculate distance and bearing for line GB

LINE BEARING DISTANCE LATIT DEPART
BC 127⁰33’36” 49.160 -29.968 38.970
CD 175⁰36’37” 52.432 -52.278 4.013
DE 235⁰34’15” 23.321 -13.185 -19.236
EF 272⁰14’20” 40.655 1.588 -40.624
FG 308⁰12’23” 44.973 27.816 -35.339
GB 66.027 52.216
check 0.000 0.000

Distance GB = √66.0272 + 52.2162
= 84.179


tan =

= tan−1 52.216
66.027

= 38°20′17"

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
Latit (N/S) Depart (E/W)
66.027 52.216 Identify the baering
position base on latit and
Bearing GB = 38°20′17" depart data.

line bearing distance latit depart
GB 38°20′17" 84.179 66.027 52.216

FIND ANGLE A

Angle A = (23⁰17’17” + 180⁰)-56⁰10’32”

= 147⁰ 06’ 45”

FIND ANGLE B

Angle B =(56⁰ 10’ 32” +180⁰) – (38⁰ 20’ 17” + 180⁰)

= 17⁰ 50’ 15”

FIND ANGLE G

Angle G = 38⁰20’17’ - 23⁰17’17”

= 15⁰ 03’ 00”

FIND DISTANCE AG USING SIN METHOD

sin 147⁰06′45" sin 17⁰50′15"
84.178 =


AG = 47.487

FIND DISTANCE AB USING SIN METHOD

sin 147⁰06′45" sin 15⁰03′01"
84.178 =


AB = 40.255

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
EXAMPLE 3
Find distance AB and bearing AG.

Calculate bearing and distance for line GB

LINE BEARING DISTANCE LATIT DEPART
BC 121⁰00’07” 39.014 -20.095 33.441
CD 72⁰10’12” 40.936 12.534 38.970
DE 181⁰09’49” 39.619 -39.611 -0.805
EF 209⁰08’52” 29.602 -25.853 -14.418
FG 295⁰23’45” 44.969 19.286 -40.623
GB 53.739 -16.565
check 0.000 0.000

Distance GB = √53.7392 + 16.5652
= 56.234


tan =

= tan−1 16.565 = 17°07′55"

53.739

Latit (N/S) Depart (E/W) Identify the bearing
53.739 -16.565 position base on latit and
depart data.
Bearing GB = 360° − 17°07′55" = 342⁰52’05”

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

Line bearing distance latit depart
GB 342⁰52’05” 56.234 53.739 -16.565

FIND ANGLE B
= (23⁰17’17” + 180⁰)-(342⁰52’05”-180⁰)
= 40⁰ 25’ 12”

FIND ANGLE A USING SIN METHOD
sin 40⁰25′12" sin

36.759 = 56.234
A = 82⁰ 42’ 12”

FIND BEARING AG
BEARING AG= 23⁰17’17’ + 82⁰ 42’ 12”

= 105⁰ 59’ 29”

FIND ANGLE G
ANGLE G = 342⁰52’05” –(105⁰59’29” +180⁰)

= 56⁰ 52’ 36”

FIND DISTANCE AB USING SIN METHOD
sin 56⁰52′36" sin 40°25′12"

= 36.759
AB = 47.480

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
SOLVE MISSING DATA FOR NON ADJUSSENT LINE
Example 4
Find distance for line FE and line CD

STEP BY STEP
1) First ,joint point for missing line and calculate bearing and distance for that line.

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
Calculate distance and bearing for line CF DEPARTMENT OF CIVIL ENGINEERING

LINE BEARING DISTANCE LATIT DEPART
FG 334⁰28’42” 52.006 46.931 -22.407
GA 309⁰40’46” 55.410 35.379 -42.645
AB 16⁰12’21” 69.926 67.148 19.516
BC 95⁰00’02” 99.401 -8.664 99.023
CF -140.794 -53.487
check
0.000 0.000

Distance CF = √140.7942 + 53.4872
= 150.611


tan =

= tan−1 53.487
140.794

= 20°48′06"

Latit (N/S) Depart (E/W)
-140.794 -53.487

Bearing CF = 180° + 20°48′06" Identify the baering
= 200⁰ 48’ 06” position base on latit and
depart data.

LINE BEARING DISTANCE LATIT DEPART
CF 200⁰ 48’ 06”
150.611 -140.794 -53.487

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

2. Draw a line parallel to line CF and start the point at another point for missing line. Draw
line parallel to FE .

3. Joint point X to point D

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

4. Calculate distance and bearing for line XD using XED triangulation

Angle XED = (227⁰ 45’ 25” - 180° )- (200° 48’ 06” - 180° )

= 26° 57’ 19”

Distance XD = √(64.439)2 + (150.611)2 − 2(64.439)(150.611) cos 26° 57’ 19”

= 97.644

sin 26° 57’ 19” sin
97.644 = 64.439

Angle EXD = 17° 24’ 23”
Bearing XD = 200°48’ 06” - 17° 24’ 23”

= 183° 23 43”

5. Find distance for line FE and line CD using DCX triangulation
Angle CXD = 250° 40’ 30” - 183° 23’ 43”
= 67° 16’ 47”
Angle XDC = 360° - (139°50’03” + 180°) + (183°23’43”-180°)
= 43° 33’ 40”
Angle XCD = 139° 50’ 03” – ( 250° 40’ 30” -180° )
= 69° 09’ 33”
sin 67° 16’ 47” sin 69° 09’ 33”
= 97.644
CD = 96.372

sin 43° 33’ 40” sin 69° 09’ 33” P R E P A R E D B Y N F Z Page 11
= 97.644

CX = 72.000 = FE

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

Example 5
1. Find distance for lineAB and bearing for line HG

STEP BY STEP
1) First ,joint point for missing line and calculate bearing and distance for that line.

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
Calculate bearing and distance for line GB DEPARTMENT OF CIVIL ENGINEERING

LINE BEARING DISTANCE LATIT DEPART
BC 108⁰36’10” 47.243 -15.071 44.775
CD 75⁰32’41” 62.322 15.557 60.349
DE 157⁰50’55” 78.736 -72.925 29.688
EF 238⁰07’11” 60.753 -32.086 -51.589
FG 284⁰57’27” 43.323 11.182 -41.855
GB 93.343 -41.368
check
0.000 0.000

Distance GB = √41.3682 + 93.3432
= 102.099


tan =

= tan−1 41.368
93.343

= 23°54′08"

Latit (N/S) Depart (E/W)
93.343 -41.368

bearing GB = 360° − 23°54′08" Identify the bearing
= 336⁰ 05’ 52” position base on latit and
depart data.

LINE BEARING DISTANCE LATIT DEPART
GB 336⁰ 05’ 52” 102.099 93.343 -41.368

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

2. Draw a line parallel to line GB and start the point at another point for missing line. Draw
line parallel to HG.

3. Joint point X to point A

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

4. Calculate distance and bearing for line XA using XAH triangulation

Angle XHA = 336° 05’ 52” - 328° 07’ 24”

= 07° 58’ 28”

XA = √(102.099)2 + (37.785)2 − 2(102.099)(37.785) cos 07° 58’ 28”

= 64.891

sin 07° 58’ 28” sin
64.891 = 37.785

Angle AXH = 04° 38’ 0”

Bearing XA = (336°05’52”-180°) + 04° 38’ 0”
= 160° 43’ 52”

5. Find distance for line AB and bearing for line HG using BXA triangulation

Angle BAX = 360°- (160° 43’ 52” + 180°) + 18° 27’ 13”

= 37° 43’ 21”

sin 37° 43’ 21” sin
44.189 = 64.891

Angle XBA = 63° 57’ 31”

Bearing BX = (18° 27’ 13” + 180°) + 63° 57’ 31”

= 262° 24’ 44” = bearing HG

Angle BXA = 160° 43’ 52” – (262° 24’ 44” - 180°)

= 78° 19’ 08”

sin 37° 43’ 21” sin 78° 19’ 08”
44.189 =


BA = 70.728

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
TUTORIAL DEPARTMENT OF CIVIL ENGINEERING

1. Calculate distance AB and AG

Answer : AB = 117.381 AG = 86.529

2. Calculate bearing for line BC and line CD

Answer : BC = 59° 28’ 57” CD= 126° 28’ 55”

b

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CHAPTER 1: MISSING LINE IN CLOSE TRAVERSE CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

3. Calculate bearing DE and distance AB

Answer Bearing DE = 160° 25’ 53” Distance AB= 94.793

4. Calculate bearing AB and EF
Answer Bearing AB = 35° 04’ 27” Bearing EF = 345° 08’ 29”

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Chapter 2 :
Boundary Problem

CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 1

Figure show a rectangular lot and the ratio of AB and BC is 3 : 2. From the data given calculate the area
of lot ABCD

STEP BY STEP
1. Rotate triangulation BAF 90°
2. Create similar shape for triangulation BA’F’ with create one line at C and connect to x
3. Calculate distance from x to B

BX = 2x = BX
BF' 3x 72.334

BX = 48.223

4. Calculate angle EBX
= 90° - 19° 53’ 07”
= 70° 06’ 53”

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

5. Calculate distance EX using cos method
= √(55.256)2 + (48.223)2 − 2(55.256)(48.223) cos 70°06′53"

= 59.716

6. Calculate angle BXE using sin method
sin 70° 06′53" sin
59.716 = 55.256
Angle B XE = 60° 28’ 30”

7. Calculate distance BC using Pythagoras theorem

sin 6028'30" = BC
48.223

BC = 41.961

8. Calculate distance BA’ and calculate area ABCD

2 41.961
′ = 3 = ′

BA’=62.942

Area = 41. 961 x 62.942
= 2641.109 unit²

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 2

Figure show a rectangular lot and the ratio of AB and BC is 3 : 2. From the data given calculate the area
of lot ABCD

STEP BY STEP

1. Rotate triangulation BAE 90°
2. Create similar shape for triangulation BA’E’ with create one line at D and connect to x
3. Calculate distance from x to A

= 2 =

′ 3 16.543

AX = 11.029

4. Calculate angle FAX

Angle FAX = 90° - 30° 37’ 03”
= 59° 22’ 57”

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

9. Calculate distance fX using cos method
= √(11.029)2 + (13.663)2 − 2(11.029)(13.663) cos 59°22′57"

= 12.443

10. Calculate angle AXF using sin method
sin 59°22′57" sin
12.443 = 13.663
AXF = 70° 54’ 10”

11. Calculate distance DA using Pythagoras theorem
sin 7054'10" = DA
11.029
DA = 10.422

12. Calculate distance BA’ and calculate area ABCD

2 10.422
′ = 3 = ′

BA’ =15.633

Area = 10.422 x 15.633
= 162.927 unit²

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 3

Figure show a rectangular lot and the ratio of AB and BC is 1 : 2. From the data given calculate the area
of lot ABCD

STEP BY STEP

1. Rotate triangulation BPC 90°
2. Create similar shape for triangulation BC’P’ with create one line at A and connect to x

3. Calculate distance BX P R E P A R E D B Y N F Z Page 23
= 1 =

′ 2 12.738

BX = 6.369
4. Calculate distance AX

1
′ = 2 = 25.610

AX = 12.805

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

5. Create line PX and calculate the distance using theorem Pythagoras
PX = √(6.369)2 + (12.738)2
= 14.242

6. Calculate angle BXP using theorem Pythagoras
12.738

tan = 6.369
X = 63° 26’ 06”

7. Calculate angle AXP using cos method
11.8492 = (12.805)2 + (14.242)2 − 2(12.805)(14.242) cos
X = 51°37’50”

8. Calculate Angle AXB
Angle BXP + Angle AXP = 63° 26’ 06” + 51°37’50”
= 115° 03’ 56”

9. Calculate distance AB
2 = (12.805)2 + (6.369)2 − 2(12.805)(6.369) cos 115° 03’ 56”

= 16.542

10. Calculate distance BC’

1 16.542
′ = 2 = ′

BC’ = 33.084

11. Area ABCD = 16.542 x 33.084
= 547.276

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

Example 4

Referring to the following figure, calculate the area for square lot of ABCD.

STEP BY STEP

1. Rotate triangulation BPA 90°
2. Distance BP = distance BP’ and distance PA = distance CP’
3. Calculate distance PP’ using theorem Pythagoras

PP’ = √(31.602)2 + (31.602)2

= 44.692
4. Calculate angle BP’P

31.602
tan ′ = 31.602

P’ = 45⁰

5. Calculate angle PP’C using cos method

cos ′ = 34.9912 + 44.6922 − 36.4502
2(34.991)(44.692)

P’ = 52° 45’00”

6. Calculate angle BP’C P R E P A R E D B Y N F Z Page 25
= angle BP’P + angle PP’ C
= 45⁰ + 52° 45’00”
=97⁰ 45’ 00”

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
7. Calculate distance BC using cos method DEPARTMENT OF CIVIL ENGINEERING

2 = (31.602)2 + (34.991)2 − 2(31.602)(34.991) cos 97⁰ 45’ 00”

= 50.212

8. Area ABCD = 50.212 x 50.212 = 2521.245 unit ²

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

Example 5

Referring to the following figure, calculate the area for square lot of ABCD.

STEP BY STEP

1. Rotate triangulation BAG 90°
2. Distance BA = distance BC and distance BC = distance BG’
3. Calculate angle FBG’

90⁰ - 43⁰ 32’ 00”
= 46⁰ 28’ 00”

4. Calculate distance FG’
( ′)2 = (75.515)2 + (77.393)2 − 2(75.515)(77.393) cos 46⁰ 28’ 00”

= 60.343

5. Calculate angle BG’F P R E P A R E D B Y N F Z Page 27
sin ′ sin 46°28′00"
77.393 = 60.343
G’ = 68⁰ 24’ 22”

6. Calculate distance BC using theorem Pythagoras


sin 68⁰ 24’ 22” = 75.515
BC = 70.215

7. Area = 70.215 x 70.215 = 4930.146 unit ²

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 6

Figure show a rectangular lot and the ratio of AB and BC is 2 : 3. From the data given calculate the area
of lot ABCD

STEP BY STEP

1. Rotate triangulation DPC 90°
2. Create similar shape for triangulation BC’P’ with create one line at A and connect to x

3. Calculate distance BX P R E P A R E D B Y N F Z Page 28
3
′ = 2 = 47.736
BX = 31.824

4. Calculate distance CX
3
′ = 2 = 35.450
CX = 23.633

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

5. Create line PX and calculate the distance using theorem Pythagoras
PX = √(35.450)2 + (23.633)2
= 42.605

6. Calculate angle CXP using theorem Pythagoras

35.450
tan = 23.633
X = 56° 18’ 37”

7. Calculate angle BXP using cos method
32.9322 = (31.824)2 + (42.605)2 − 2(31.824)(42.605) cos
X = 49° 59’ 23”

8. Calculate Angle CXB
Angle CXP + Angle BXP = 56° 18’ 37”+ 49° 59’ 23”
= 106° 18’ 00”

9. Calculate distance AB
2 = (31.824)2 + (23.633)2 − 2(31.824)(23.633) cos 106° 18’ 00”

= 44.648

10. Calculate distance BC’

2 44.648
′ = 3 = ′

BC’ = 66.972

11. Area ABCD = 44.648x 66.972
= 2990.166

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 7

Figure show 2 lot with boundaries ABCD. Both lot have same area. Landlords agree to change the
boundaries to straight line and maintained the area. The boundaries must start from A.

Step by step ;
1. First, draw straight between A to D and calculate bearing and distance for that line.

LINE BEARING DISTANCE LATIT DEPART
DC 31⁰ 17’ 28” 48.568 41.503 25.226
CB 313⁰28’53” 40.389 27.792 -29.306
BA 27⁰49’37” 46.092 40.762 21.516
AD -110.057 -17.436
check
0.000 0.000

Distance AD = √110.0572 + 17.4362
= 111.430

tan = = tan−1 17.436
110.057

= 9° 0′09"

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
Latit (N/S) Depart (E/W)
-110.057 -17.436 Identify the baering
position base on latit and
Bearing AD = 180° − 9° 0′09" depart data.
= 189° 00’ 09”

LINE BEARING DISTANCE LATIT DEPART
AD 189° 00’ 09”
111.430 -110.057 -17.436

2. Calculate distance from AX and BX using sin method

Angle BAX = 207°49’ 37” - 189° 0’09”

= 18° 49’ 28”

Angle XBA =133°28’53”- (207° 49’37” -180°)

= 105° 39’ 16”

Angle AXB =360°- (133° 28’53”+180⁰ ) + (189°0’9” -180°)

= 55° 31’16”

Distance BX

sin 18° 49’ 28” sin 55° 31’16”
= 46.092

BX = 18.042

Distance AX
sin 105° 39’ 16” sin 55° 31’16”

= 46.092
AX = 53.840

3. Calculate area for triangulation ABX using sin method

Area = ( )( ) sin

2

(46.092)(53.840) sin 18° 49’ 28”
=2

= 400.367

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
4. Calculate distance CX and DX
CX= 40.389 - 18.042 = 22.347
DX= 111.430 - 53.840 = 57.590

5. Calculate area for triangulation DCX

Angle DXC = Angle AXB

= 55° 31’16”

Area ( )( ) sin
=2

(22.347)(57.590) sin 55° 31’16”
=2

= 530.445

6. Calculate different area between triangulation DCX and ABX

Different area = 530.445- 400.367

= 130.078

7. Area DCX > area ABX, so move line AD to DCX
8. Calculate bearing and distance DD’

Angle ADD’ = 82°55’27”- (189°0’9”-180°)

= 73° 55’18”

= ( )( ′) sin 73° 55’18”

22

1302.078= (111.430)( ′) sin 73° 55’18”
2

DD’ = 1.215

AD’ = √(111.430)2 + (1.215)2 − 2(111.430)(1.215) cos 73°55′18"
= 111.100

sin sin 73° 55’18"
1.215 = 111.100
Angle A = 0° 36’ 08”
Bearing AD’ = 189° 0’9” - 0° 36’ 08” = 188° 24’ 01”

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 8

Figure show 2 lot with boundaries ABCDE. Both lot have same area. Landlords agree to change the
boundaries to straight line and maintained the area. The boundaries must start from E.

Step by step ;
1. First, draw straight between A to E and calculate bearing and distance for that line.

LINE BEARING DISTANCE LATIT DEPART
AB 135⁰ 38’ 7” 33.142 -23.693 23.174
BC 168⁰55’5” 25.531 -25.055 4.907
CD 247⁰39’43” 41.560 -15.796 -38.441
DE 140°29’37” 51.627 -39.833 32.843
EA 104.377 -22.483
check
0.000 0.000

Distance EA = √104.3772 + 22.4832
= 106.771

tan = = tan−1 22.483
104.377

= 12°09′21"

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
Latit (N/S) Depart (E/W)
104.377 -22.483 Identify the baering
position base on latit and
Bearing EA = 360° − 12°09′21" depart data.
= 347° 50’ 39”

line bearing distance latit depart
EA 347° 50’ 39” 106.771 104.377 -22.483

2. calculate distance from DX and EX using sin method
Angle EDX = 140° 29’37” -(247°39’ 43”-180°)
= 72° 49’ 54”
Angle DXE =247°39’43”-( 347°50’39”-180°)
= 79° 49’ 04”
Angle XED =347° 50’39” - (140°29’37” +180°)
= 27° 21’02”
Distance DX
sin 27° 21’02” sin 79° 49’ 04”
= 51.627
DX = 24.099

Distance EX
sin 72° 49’ 54” sin 79° 49’ 04”

= 51.627
EX = 50.116

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

3. Calculate area for triangulation DXE using sin method

Area = ( )( ) sin

2

(50.116)(24.099) sin 79° 49’ 04”
=2

= 594.362

4. Calculate distance AX and CX
CX = 41.560 - 24.099 = 17.461
AX = 106.771 - 50.116 = 56.655

5. Calculate area for ABCX- (you can use any method)

Angle ABC = (135°38’07’+180°)-168⁰55’5”

= 146° 43’ 02”

Area ABC =(33.142 )(25.531) sin 146° 43’ 02”
2

= 232.171

Angle AXC = Angle DXE

= 79° 49’ 04”

Area AXC = (56.655)(17.461) sin 79° 49’ 04”

2

= 486.836

Area for ABCX = area AXC + Arae ABC
= 486.836 + 232.171
= 719.007

6. Calculate different area between triangulation DXE and ABCX

Different area = 719.007- 594.362

= 124.645

7. Area ABCX > area DEX, so move line AE to ABCX

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

8. Calculate bearing and distance AA’

Angle EAA’ = (347° 50’ 39” 180°)- 87°42’13”

= 80° 08’26”

= ( )( ′) sin 73° 55’20”
2 2

124.645 (106.771)( ′) sin 80° 08’26”
2 =2

AA’ =1.185

9. Calculate distance for line EA’
EA’ = √(106.771)2 + (1.185)2 − 2(106.771)(1.185) cos 80° 08’26”
= 106.574

sin sin 80° 08’26”
1.185 = 106.574
Angle E = 0° 37’ 40”

Bearing EA’ = 347° 50’ 39” + 0° 37’ 40”
= 348° 28’ 19”

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

TUTORIAL

1. Figure show a rectangular lot and the ratio of AB and BC is 1 : 3. From the data given calculate
the area of lot ABCD

Answer : AB = 32.410 BC= 97.230 Area = 3151.224

P

2. Figure show a rectangular lot and the ratio of AB and BC is 3 : 2. From the data given calculate
the area of lot ABCD
Answer: AB = 138.693 BC = 92.462 area= 12823.832

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

3. Figure show 2 lot with boundaries ABCD. Both lot have same area. Landlords agree to change
the boundaries to straight line and maintained the area. The boundaries must start from D.

Answer : bearing and distance AD = 149.091 171° 51’ 45”
Bearing and distance AA’ = 0.758 71° 17’ 59”
352°08’ 55”
: bearing and distance DA’ = 149.232

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CHAPTER 2: BOUNDARY PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
BC : BA = 1: 2
Answer BC = 82.321 BA = 164.642 AC: CD = 1: 1
Answer BA = 80.123

BA: BC = 1:3 BA: BD = 1:1
Answer BA =229.563 BC= 76.521 Answer BA = 112.000

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Chapter 3:
Area Division

CHAPTER 3: AREA DIVISION CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 1

The area for Lot230 is 8330.499m². Land lord want to apply subdivision for the lot where the area for lot
ABXY is 1/3 from the whole area. Distance from D to Y is 44.351m. Calculate distance and bearing for
line XY.

LOT 230

Step by step

1. Calculate area for lot ABXY

area = 1  8330.499 = 2776.833
3

2. Draw line from x to y
3. Connect line A to Y (distance D to Y was given)
4. Calculate bearing and distance for line AY

LINE BEARING DISTANCE LATIT DEPART
YD 278°36’26” 44.351 6.638 -43.851
DA 15°20’52” 54.868 52.911 14.522
AY 153° 46’ 44” 66.380 -59.549 29.329

Bearing AY = 180° - 26° 13’ 16”
= 153° 46’ 44”

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CHAPTER 3: AREA DIVISION CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

5. Calculate area for triangulation ADY

Area = 1  ( AD)(DY ) sin D
Angle D 2

= (278°36’26”-180°) - 15°20’52”

= 83° 15’ 34”

areaADY = 1  (54.868)(44.351) sin 8315'34"
2

= 1208.315

6. Calculate area for triangulation AXY

Area AXY = Area AXYD – area ADY

= 2776.833 – 1208.315

= 1568.518

7. Calculate distance AX using area formula

Calculate angle A = 153° 46’ 44” - 88° 57’ 42”
= 64°49’02”

area = 1  ( AX )( AY) sin A
2

1568.518 = 1  (AX )(66.380) sin 6449'02"
2

AX =52.222

8. Calculate bearing and distance for line XY

LINE BEARING DISTANCE LATIT DEPART

YA 333° 46’ 44” 66.380 59.549 -29.329
AX 88°57’42” 52.222 0.946 52.213
XY 200° 43’ 14” 64.679 -60.495 -22.884

Bearing XY = 180° + 20° 43’ 14”
= 200° 43’ 14”

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CHAPTER 3: AREA DIVISION CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 2

The area for Lot 252 is 27047.750 m². Land lord want to apply subdivision for the lot where the area for
lot ABXE is 1/2 from the whole area. Calculate distance and bearing for line BX.

LOT 252

Step by step

1. Calculate area for lot ABXE

area = 1  27047.750 = 13523.875
2

2. Draw line from B to X
3. Connect line E to B (distance D to B was given)
4. Calculate bearing and distance for line BE

LINE BEARING DISTANCE LATIT DEPART

EA 350°28’44” 96.903 95.568 -16.029
AB 77°01’08” 128.996 28.976 125.699
BE 221° 21’ 59” 165.948 -124.544 -109.670

Bearing BE = 180° + 41° 21’ 59”
= 221° 21’ 59”

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