SURVEY COMPUTATION 2

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

5. Calculate area for triangulation ABE

area = 1  ( AE)( AB) sin A
Angle D 2

= (350°28’44”-180°) - 77°01’08”

= 93°27’36”

areaABE = 1  (96.903)(128.996) sin 9327'36"
2

= 6238.657

6. Calculate area for triangulation XEB

Area AXY = Area ABXE – area ABE

= 13523.875 – 6238.657

= 7285.218

7. Calculate distance E X using area formula

Calculate angle A = (271° 37’ 12”-180°) – (221° 21’ 59” -180°)
= 50° 15’ 13”

area = 1  (EB)(EX ) sin C
2

7285.218 = 1  (EX )(165.948) sin 50°15'13"
2

EX =114.193

8. Calculate bearing and distance for line BX

LINE BEARING DISTANCE LATIT DEPART

XE 271° 37’ 12” 114.193 3.228 -114.147
EB 41° 21’ 59” 165.948 124.544 109.670
BX 177° 59’42” 127.850 -127.772 4.477

Bearing AY = 180° - 2° 0’ 24”
= 177° 59’36”

P R E P A R E D B Y N F Z Page 44

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

The area for Lot 301 is 3299.146 m². Land lord want to apply subdivision for the lot where the area for
lot AXYE is 1113.982 m².Given bearing for line XY is 175° 54’ 34”. Calculate distance for line AX,XY and
line YE.

LOT 301

Step by step:

1. Draw line XY
2. Project line XY to AP
3. Calculate distance AP using sin method

Angle AEP = (277°19’40”-180°)-10°11’16”

= 87°08’24”

Angle APE =(175°54’34” +180°)-277°19’40”

= 78°34’54”

sin 7834'54" = sin 8708'24"
33.810 AP

AP =34.450

P R E P A R E D B Y N F Z Page 45

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

4. Calculate area AEP
Angle EAP = (10°11’16”+180°)- 175°54’34”
= 14°16’42”

Area = 1 (AE)( AP)sin A
2

= 1 (33.810)(34.450)sin1416'42"
2

=143.633

5. Calculate area AXYP

Area AXYP = area AXYE – area APE

= 1113.982 - 143.633

= 970.349

6. Draw line perpendicular to line AP and line XY (label as d)
7. From Pythagoras formula:-

Angle MAX = 175°54’34”-89°42’50”
= 86°11’44”

Tan 86°11’44” = d
AM

AM = d
tan 8611'44"

AM = d kot 86°11’44”

Angle NYP =(175°54’34”+180°)-277°19’40”
= 78°34’54”
d = d  kot 8611'44"
tan 8611'44"

NY = d kot 78°34’54”

P R E P A R E D B Y N F Z Page 46

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

8. Calculate Area AXYP
Area AXYP = area AMX + area MXNP + area PNY

Area AMX = 1  (AM )d = 1  (dkot8611'44")d = 1  d 2kot8611'44"
22 2

= 0.033d²

Area PNY = 1  d 2kot7834'54"
2

= 0.101d²

Area MXNP = d(AP-MA)

= d(34.450-d kot 86°11’44”)

= 34.450d-0.066d²

970.349 = 0.033d² + 34.450d-0.066d² +0.101d²

970.349 = 0.068d²+34.450d

Convert to quadratic equation

0.068d² + 34.450d - 970.349 = 0 calculate value d using calculator

d = 26.750 ,-53.337 (take positive value only)

9. Calculate distance XY
XY = XN +NY
XN =MP
MP =AP-AM

XY = (AP-AM)+NY
= (34.450 -d kot 86°11’44”)+d kot 78°34’54”
= (34.450 –26.750 kot 86°11’44”)+26.750 kot 78°34’54”
=38.074

P R E P A R E D B Y N F Z Page 47

CHAPTER 3: AREA DIVISION CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
10. Calculate AX and EY using trigonometry

sin 86°11’44” = 26.750
AX

AX = 26.809

sin 78°34’54” = 26.750
PY

PY = 27.290

sin1416'42" = sin 7834'54"
EP 33.810

EP = 8.507
EY = PY + EP

= 27.290 + 8.507
= 35.797

P R E P A R E D B Y N F Z Page 48

CHAPTER 3: AREA DIVISION CG 203 SURVEY COMPUTATION 2
TUTORIAL 1 DEPARTMENT OF CIVIL ENGINEERING

The area for Lot243 is 31072.645m². Land lord want to apply subdivision for the lot where the

area for lot ABXY is 1/3 from the whole area. Distance from B to X is 58.310 m. Calculate

distance and bearing for line XY.

Answer : bearing and distance XY = 167.977 183° 30’ 26”

TUTORIAL 2

The area for Lot 3241 is 35032.567m². Land lord want to apply subdivision for the lot where the
area for lot ABCXY is 18396.070².Given bearing for line XY is 352° 43’ 04”. Calculate distance

for line XY

Answer : bearing and distance XY = 167.977 183° 30’ 26”

P R E P A R E D B Y N F Z Page 49

Chapter 4 : Route
Problems

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
PROVE:-Road with different width

PROVE:-Road with different width

= −


Step by step

1. Solve triangulation ABE, BEC and ADC

2. Solve triangulation ABE

tan = 2

2
= tan = 2 cot

P R E P A R E D B Y N F Z Page 51

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

3. Solve triangulation BEC

tan = 2


= 2 = 2 cot
tan

4. Solve triangulation ADC

sin = 1


= 1 = 1 cosec
sin

= +

+ = 1 cosec

2 cot + 2 cot = 1

2( cot + cot ) = 1

( cot + cot ) = 1
2

cot = 1 − cot …… prove

2

P R E P A R E D B Y N F Z Page 52

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 1 : SAME WIDTH

Calculate bearing and distance for road secant AB. Given the width of road is 30 m.

Step by step

1. Joint point A to B
2. Calculate angle B = (360° - 294° 17’ 52”) + 56° 02’ 05”

= 121°44’13”

3. Calculate α = angleB P
2

α = 121°44'13"
2

` = 60° 52’ 7”

4. Calculate bearing BA = 294°17’52” + 60° 52’ 7”
= 355° 09’ 59”

5. Create line A to P perpendicular to road and distance for line AP equal to
Width of the road

6. Calculate distance AB using theorem’s Pythagoras

sin 60° 52’ 07” = 30
AB

AB = 34.344

P R E P A R E D B Y N F Z Page 53

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
EXAMPLE 2: DIFFERENT WIDTH
Calculate bearing and distance for road secant AB

20.0 40.0
00 00

Step by step
1. Joint point A to B
2. Create line BP perpendicular to road, project line from left road.
3. Calculate angle θ

117° 18’ 53” - (230° 24’ 46” -180°) = 66° 54’ 07”

P R E P A R E D B Y N F Z Page 54

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
4. Calculate angle cot α = 20 cosec − cot
40 1
cot = tan
cot α = 20 cosec 66° 54' 07" − cot66° 54' 07"
40 1
= sin
20 1 
1=  sin 66° 54' 07"  −  1 
tan  40  tan 66° 54' 07" 

1 = 0.11708
tan 

1 = tan 
0.11708

α = 83° 19’ 20”

5. Calculate bearing AB = 117° 18’ 53” + 83° 19’ 20”
= 200° 38 13”

7. Calculate distance AB using theorem’s Pythagoras

sin 83° 19’ 20” = 40
AB AB

= 40.273

P R E P A R E D B Y N F Z Page 55

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 3

Calculate distance CB and area ABCD. Given the width of road is 25m

Step by step

1. Draw line CP and line BQ perpendicular to line DA.
2. Distance CP and BQ = width of road =25
3. Calculate distance DP and QA using theorem’s Pythagoras

Calculate angle CDP = (22° 13’ 37” + 180°) –(319° 51’14” -180°)
= 62° 22’ 23”

Tan 62° 22’ 23” = = 25


DP = 13.085

Calculate angle CDP = 319° 51’14” –(89° 02’ 56”+ 180°)
= 50° 48’ 18”

Tan 50° 48’ 18” = = 25


QA = 20.386

4. Distance CB = 65.094 -20.386 – 13.085 = 31.623
5. Area ABCD
= 1 (65.094 + 31.623) × 25
2

= 1208.963

P R E P A R E D B Y N F Z Page 56

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
EXAMPLE 4
Calculate distance for secant BD and width of road AE.

Step by step

1. Create line D to P and this line perpendicular to the road

2. Calculate angle B = (8° 13’ 43” + 180°) - 110° 35’ 20”

= 77° 38’ 23”

3. Distance BD = 25
Sin 77° 38’ 23” BD

BD = 25.593

4. Calculate angle CDB = 110°35’20” - 8° 13’ 43”

= 10221'37"

5. calculate bearing BC using sin method

sin10221'37" = sin
66.232 25.593

θ = 22° 10’ 35”

Bearing BC =( 110° 35’ 20” + 180°) + 22° 10’ 35”
= 312° 45’ 55”

P R E P A R E D B Y N F Z Page 57

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

6. Calculate bearing DE using latit depart

LINE BEARING DISTANCE LATIT DEPART
DB 08⁰13’43” 25.593 25.330 3.663
DC 132⁰45’55” 66.232 -44.971 48.624
CE 274⁰36’36” 75.665 6.081 -75.420

ED ° ′ " 26.814 13.560 23.133

Distance CE = √13.5602 + 23.1332
= 26.814

tan
=

= tan−1 23.133
13.560

= 59° 37′20"

Bearing ED = Bearing AE

7. Calculate θ = 110°35’20” - 59° 37′20"
= 50° 58’ 00”

8. Calculate width of road AE using formula

cot77° 38’ 23” = AE cosec50° 58' 00" − cot50° 58' 00"
AE 25

= 25 (cot 77° 38’ 23”+ cot50° 58' 00" )
cosec50° 58' 00"

=20.000

P R E P A R E D B Y N F Z Page 58

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
EXAMPLE 5
Calculate bearing and distance for CB.

Step by step

1. Joint line AB
2. Draw line AP perpendicular to road.
3. Project left road to calculate angle for θ.
4. Calculate angle θ = (292° 59’ 26” - 180°) - 51° 58’ 25”

= 61° 01’ 01”
5. Calculate angle α using formula

cotα = 40 cosec61° 01' 01" − cot61° 01' 01"
30

40 1 
1=  sin 61° 01' 01"  −  1 
tan  30  tan 61° 01' 01" 

1 = 0.97029978
tan 

1 = tan
0.97029978

α = 45° 51’ 49”

P R E P A R E D B Y N F Z Page 59

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
6. Calculate baring BA = 51° 58’ 25” - 45° 51’ 49”
= 06° 06’ 36”

7. Calculate distance BA using theorem’s Pythagoras

Sin 45° 51’ 49” = AP = 30
AB AB

= 41.801

8. Calculate bearing and distance BC using latit and depart

LINE BEARING DISTANCE LATIT DEPART
BA 06⁰06’36” 41.801 41.564 4.449
AC 323⁰54’27” 93.068 75.205 -54.825

CB 156°39’50” 127.172 -116.769 50.376

Distance CB = √116.7692 + 50.3762
= 127.172

tan
=

= tan−1 50.376
116.769

= 23° 20′10"

Bearing CB = 180°-23° 20′10"
= 156° 39’ 50”

P R E P A R E D B Y N F Z Page 60

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
EXAMPLE 6
Given width of road is 25 m. calculate secant BA

Step by step :-

1. Calculate angle CAD using cos method
59.446² = 38.651² + 30.612² - 2(38.651)(30.612 cos θ
θ = 117°46’40

2. Calculate bearing DA
Calculate angle CDA using sin method

sin117°46'40 = sin D
59.446 38.651

D = 35° 07’ 04”
Bearing DA = (97° 09’ 05” +180°) - 35° 07’ 04”
= 242° 02’ 01”

3. Calculate angle BAD

117°46'40 = 5853'20"
2

4. Calculate bearing AB

(242° 02’ 01”- 180°) - 5853'20"

= 03° 08’ 41”

5. Create line BP perpendicular to road

6. Calculate distance AB using theorem’s Pythagoras

sin 5853'20"= 25
AB

AB = 29.200

P R E P A R E D B Y N F Z Page 61

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
EXAMPLE 7

Calculate bearing and distance for secant BE

40.000

40.000

Step by step

Calculate Angle ABC

(45.380+59.983)² = 53.279² + 70.425² - 2(53.279)(70.425) cos θ

θ = 116° 06’48”

Angle EBC = 116° 06'48"
2

= 58°03’ 24”

Calculate Distance BE

Sin 58°03’ 24” = 40
BE BE

= 47.138

sin  = sin 11606'48"
53.279 105.363

θ = 27° 00’ 15”
= (89° 38’ 36” +180°) + 27° 00’ 15”
bearing BC = 296° 38’ 51”
= (296° 38’ 51”-180°) + 58°03’ 24”
Bearing BE = 174° 42’ 15”

P R E P A R E D B Y N F Z Page 62

CHAPTER 4: ROUTE PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

TUTORIAL 1

Calculate secant AB, distance CB and area ABCD .Given the width of road is 25m

answer : AB = 32.258 CB = 31.623 area= 1208.963

TUTORIAL 2
Calculate secant AB and secant CD, after that calculate area for road ABCD.
Answer BA= 13° 15’ 41” , 33.648 DC = 352° 13’ 36” , 30.159 area ABCD = 2571.775m²

P R E P A R E D B Y N F Z Page 63

Chapter 5 :

Three Point & Three
Distance Problems

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
THREE POINT & THREE DISTANCES PROBLEMS DEPARTMENT OF CIVIL ENGINEERING
EXAMPLE 1
Calculate distance for line BP

Step by step Janganabaikan
1. Calculate angle ABC using cos method value negative

109.232² = 65.004² + 77.683² - 2(65.004)(77.683) cos θ
θ = 99° 31’ 35”

2. Calculate angle φ using formula Distance a < distance b
Tan φ = a sin q
b sin p

Tan φ = 65.004 sin 2803'15"
77.683sin 2105'17"

Φ = 47° 33’ 53”

3. Calculate angle X+Y = 360 –( 2803'15"+ 2105'17" )-99° 31’ 35”

= 211° 19’ 53”

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CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
4. Calculate angle x-y using formula DEPARTMENT OF CIVIL ENGINEERING

tan x − y = tan( − 45) tan (x + y)
22
= tan(47° 33' 53"−45) tan 211° 19' 53"
2

x − y = −0904' 29"
2

x-y = -18° 08’ 58”
5. Calculate value of x and y using simultaneous equation method

x + y = 211° 19’ 53” Reminder :
x - y = -18° 08’ 58”
Position of angle x must at
2y = 229° 28’ 51” triangulation with line b

Y = 114° 44’ 26” Distance of b > distance a

x + 114° 44’ 26” = 211° 19’ 53”
x = 96° 35’ 27”

6. Calculate distance for line BP using sin method

sin114° 44' 26" = sin 2105'17"
BP 65.004

BP = 164.083

Reminder :

Use angle y to calculate BP
because small distance will
give better result

P R E P A R E D B Y N F Z Page 66

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
EXAMPLE 2 DEPARTMENT OF CIVIL ENGINEERING
Calculate distance for BP

Step by step

1. Calculate angle ABC

= (329° 37’ 10” -180°) -43° 45’ 09”
= 105° 52’ 01”

2. Calculate angle φ using formula

Tan φ = a sin q Distance a < distance b
b sin p

Tan φ = 56.694 sin 2907'19"
79.145 sin 2001'13"

Φ = 45° 31’ 10”

3. Calculate angle X+Y = 360 –( 2907'19" + 2001'13" )-105° 52’ 01”

= 204° 59’ 27”

P R E P A R E D B Y N F Z Page 67

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

4. Calculate angle x-y using formula

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

= tan(45° 31'10"−45) tan 204° 59' 27"
2

x − y = −0220' 34"
2

x-y = -04° 41’ 07”
5. Calculate value of x and y using simultaneous equation method

x + y = 204° 59’ 27”
x - y = -04° 41’ 07”

2y = 209° 40’ 34”

Y = 104° 50’ 17”

x + 104° 50’ 17” = 204° 59’ 27”
x = 100° 09’ 10”

6. Calculate distance for line BP using sin method

sin104° 50'17" = sin 2001'13"
BP 56.694

BP = 160.079

P R E P A R E D B Y N F Z Page 68

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
EXAMPLE 3 DEPARTMENT OF CIVIL ENGINEERING
Calculate distance PB, PC and PA

Step by step:-

1. Calculate angle x + y = 360° - (117°07’17” + 116°23’47”) - 62°10’10”
= 64° 18’ 46”

2. Calculate angle φ using formula

Tan φ = a sin q Distance a < distance b
b sin p

Tan φ = 103.1sin11623'47"
122.504 sin11707'17"

Φ = 40° 15’ 52”

3. Calculate angle x-y using formula

tan x − y = tan( − 45) tan (x + y)
22
= tan(40°15' 52"−45) tan 64°18' 46"
2

x − y = −0557' 45"

P R E P A R E D B Y N F Z Page 69

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

7. Calculate value of x and y using simultaneous equation method

x + y = 64° 18’ 46”

x − y = −0557' 45"

2y = 70° 16’ 31”

Y = 35° 08’ 16”

x + 35° 08’ 16” = 64° 18’ 46”
x = 29° 10’ 30”

8. Calculate distance for line BP,PA & PC using sin method

sin 35° 08'16" = sin11707'17"
BP 103.100

BP = 66.668

Angle ABP = 180° - (117°07’17” + 35° 08’ 16”)
= 27° 44’ 27”

sin 27° 44' 27" = sin11707'17"
AP 103.100

AP = 53.919

Angle PBC = 62° 10’ 10” - 27° 44’ 27”
= 34° 25’ 43”

sin 34° 25' 43" = sin11623'47"
PC 122.504

BP = 77.323

P R E P A R E D B Y N F Z Page 70

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
EXAMPLE 4 DEPARTMENT OF CIVIL ENGINEERING
Calculate distance for line BC

BC(AB + BC + CD) = sin Q • sin( P + Q + R)
( AB)(CD) (sin P)(sin R)

BC(35.716 + BC + 35.588) = sin 2703'50"•sin(1619'25"+2703'50"+1437'31')
(35.716)(35.588) (sin 1619'25" )(sin 1437'31' )

BC(71.304 + BC) = 0.45498 • sin(5800'46")
1271.061 (0.28106)(0.25250)

BC2 + 71.304BC = 0.3859
1271.061 0.07096

BC² +71.304BC = 6912.379

Solve value of BC using quadratic formula Calculate using
calculator
BC² +71.304BC - 6912.379 = 0
Select positive
BC = 54. 810 value only

P R E P A R E D B Y N F Z Page 71

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING
EXAMPLE 5
Calculate distance for line BC

STEP BY STEP

1. Calculate distance AB using Theorems Pythagoras

150 – 149.830 = 0.170

250 – 282.424 = 32.424 Ignore

negative value

Distance AB = (0.170)2 + (32.424)2

AB = 32.424

2. Calculate angle APB, PBC and CPD
Angle APB = 352°59’16” - 336° 04’ 27” = 16°54’49”
Angle BPC = (360° - = 352°59’16”) + 15°44’07”= 22°44’51”
Angle CPD = 37°16’23” - 15°44’07”= 21°32’16”

3. Calculate distance BC using formula

P R E P A R E D B Y N F Z Page 72

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

BC(32.424 + BC + 48.767) = sin 2244'51"•sin(1654'49"+2244'51"+2132'16')
(32.424)(48.767) (sin1654'49")(sin 2132'16')

BC(81.191 + BC) = 0.38667 • sin(6111'56")
1581.221 (0.290929 )(0.367115 )

BC2 + 81.191BC = 0.338838
1581.221 0.106804

BC² +81.191BC = 5016.457822

Solve value of BC using quadratic formula Calculate using
calculator
BC² +81.191BC – 5016.457822 = 0
Select positive
BC = 41.041 value only

P R E P A R E D B Y N F Z Page 73

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
DEPARTMENT OF CIVIL ENGINEERING

EXAMPLE 6

Calculate distance PB, PC and PA

Given :-

Distance B-A = 1321.06 m

Distance A-C = 1376.68 m

α = 890 15’ 30”
β = 1280 20’ 10”

angle BAC = 1020 45’ 20”

Calculate distance PB, PC and PA

Step by step

Find equation for x + y and x - y
Angle x + y =360° - (890 15’ 30”+1280 20’ 10”) -1020 45’ 20”

= 39° 39’ 00”

Tan φ = 1321.06 sin 12820'10"
1376.68sin 8915'30"

φ= 36° 58’ 15”

tan x − y = tan (36° 58’ 15” - 45°) .tan 39° 39' 00"
22

x- y = -5° 49’ 21”

P R E P A R E D B Y N F Z Page 74

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2
Find value x and y DEPARTMENT OF CIVIL ENGINEERING
x + y = 39° 39’ 00”
x- y = -5° 49’ 21”
2y = 45° 28’ 21”
Y =22° 44’ 11”
x + 22° 44’ 11” = 39° 39’ 00”

x = 16°54’49”

Calculate distance PA
sin 16°54'49" = sin 12820'10"

PA 1376.68
PA = 510.612

Calculate Distance PC
Angle PAC = 180° -(16°54’49”+1280 20’ 10”)

= 34° 45’ 01”
sin 34° 45' 01" = sin 12820'10"

PC 1376.68
PC =1000.412

Calculate Distance PB
Angle BAP = 102° 45’ 20” - 34° 45’ 01”

= 68° 0’19”
sin 68° 0'19" = sin 8915'30"

PB 1321.06
PB = 1225.158

P R E P A R E D B Y N F Z Page 75

CHAPTER 5: THREE POINT & THREE DISTANCES PROBLEMS CG 203 SURVEY COMPUTATION 2

DEPARTMENT OF CIVIL ENGINEERING

TUTORIAL 1 TUTORIAL 2

Calculate distance for line BC Calculate distance for line BP
Answer BC = 38.528 Answer BP = 107.603

TUTORIAL 3 TUTORIAL 4

Calculate distance AP, PB and PC Calculate distance AP, PB and PC

Answer: AP = 41.381 PB = 37.555 PC = 70.550 Answer: AP = 29.237 PB = 27.838 PC = 39.062

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