Example Problems in Higher Surveying
Example: Missing Data Case 6
Determine the bearings of the non-adjacent sides. The other description of the lot is given
below
Line
AB
BC
CD
DE
EA
Bearing
S 72°25’ E
S 47°05’ W
N 1°30’ W
Length (m)
12.20
12.45
13.70
14.55
10.52
Solution:
A
Plot the traverse
12.20m
Construct parallelogram, BCDD’, where
Line BC = Line D’D and Line CD = Line BD’
72°25’
EA
AB
BD’= CD
? D’E
Bearing
N 1°30’ W
S 72°25’ E
S 47°05’ W
Length
(m)
10.52
12.20
13.70
Length D’E = √ (2.5)2 + 1.32)2
12.45m
1°30’
Consider polygon, ABD’EA
Line
B
10.52m
E
Latitude
Departure
+ 10.52
- 3.69
- 9.33
(-) + 2.5
0.00
- 0.28
+ 11.63 14.55m
- 10.03
(+) – 1.32
0.00
C
D’
13.70m
47°05’
D
= 2.83m
Tan (Bearing D’E) = dep/lat = 1.32 / 2.50
Bearing D’E = N 27° 50.04’ W
First Approach: (See Figure A)
Consider ∆ DD’E and adjust figure as shown
E φ
2.83m
θ
D’
14.55m
By Cosine law
2
2
2
2.83 = 14.55 + 12.45 – 2(14.55)(12.45)cos α
α = 8°4.96’
12.452 = 2.832 + 14.552 – 2(2.83)(14.55)cos φ
φ = 38°12.59’
Higher Surveying Notes of AM Fillone, DLSU-Manila
12.45m
Figure A
α
D
Example Problems in Higher Surveying
therefore, θ = 180° - (α + φ) = 133°42.45’
Bearing BC = Bearing D’D = 180 - 133°42.45’ + 27° 50.04’
= S 74°7.59’ E
ANS.
Bearing DE = N (74°7.59’ - 8°4.96’) W = N 66°2.63’ W
ANS.
Another approach: Figure B
Consider ∆ DD’E and adjust figure as shown
E
Bearing BC = Bearing D’D = 180 – (27° 50.04’ + 133°42.45’)
= S 18°27.51’ W
ANS.
Bearing DE = N (38°12.59’ - 27°50.04’) E
= N 10°22.55’ E
ANS.
φ
14.55m
α
D
Higher Surveying Notes of AM Fillone, DLSU-Manila
27° 50.04’
2.83m
θ
D’
12.45m
Figure B