IMPORTANT FORMULA.
3. Distance = 2 2 ) ()( depart latit
4. Angle = tan-1 depart
latit
Triangulation
1. c² = a² + b² - 2ab cos C
2. area = C absin 2 1
3. area = ) )()(( c sbsass
s= 2 a+b+c
2
4. sin A = sin B
a b
Road secant
a = bearing BA - bearing BC
2
θ = bearing AD – bearing AE
cot α = L1 cosesθ -cot θ
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 (Φ - 45o) tan ( x+y )
CHAPTER 1 MISSING DATA FOR ADJUSSENT LINE
EXAMPLE 1
1) Find bearing for line CD and line DE.
1) First ,joint point from C to E, after that, calculate bearing and distance for that line.
Calculate bearing and distance CE
LINE BEARING DISTANCE LATIT DEPART
EF 281:36’23” 61.448 12.363 -60.192
FG 353:39’11” 53.430 53.103 -5.907
GA 50:05’59” 49.496 31.749 37.971
AB 94:17’04” 45.129 -3.371 45.003
BC 127:33’36” 49.160 -29.968 38.970
CE -63.876 -55.845
check 0.000 0.000
Latit = distance x cos θ
Depart = distance x sin θ
Calculate total of latit and depart
Distance CE = √ 63.876 ^2 +550845^2
= 84.846
tan θ = depart
latit
θ = tan^-1 55.845
63.876
θ = 41.0944
Latit (N/S) = -63.876
Depart (E/W) = -55.845
Identify the bearing position base on latit and depart data.
Bearing CE = 180: + 41.09'44"
= 221: 09’ 44”
LINE BEARING DISTANCE LATIT DEPART
CE 221: 09’ 44” 84.846 -63.876 -55.845
i. calculate angle of c using cos method
ED^2=EC^2+CD^2-(EC)(CD)cos C
60.793^2=84.846^2+52.432^2 - 2(84.432)(52.432) cos C
C=45.21'20"
ii. bearing CD = 221.09’ 44” - 45: 21’ 20”
= 175: 48’ 24”
iii. Calculate bearing DE using sin method.
sin 45.21'20" = sin θ
60.793 84.846
θ = 83.12'28"
So, bearing DE = (180 + 175: 48’ 24”) - 96: 47’ 32
= 259.00'52"
EXAMPLE 2
I) Find bearing for line AG and line AB
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