1. A satellite station S was set up 12.10 m from station C inside triangle ABC, as C was a church spire and could not be occupied. 2. The angle CSA was measured to be 135°40'30" and angle ASB was 71°29'30". 3. Lengths AC and BC were 2511 m and 1894 m respectively. 4. Using the formula that relates the satellite station offset to the measured angles, the corrections for angles CSA and ASB were calculated. 5. The corrected angles were then used to calculate the angle ACB according to the formula for angles of a triangle.

1. Lecture – 07
Adjustment of Angles

2. Adjustment of Angles
 After completion of field work of measurements of angles, it
is necessary to adjust the angles. Generally the angles of a
triangle and chain of triangles are adjusted under two heads.
1. Station Adjustment
2. Figure Adjustment

3. 1. Station Adjustment
 Sum of the angles about a station should be 360o
. If not, find
the difference and adjust the difference equally to all the
angles algebraically to make their sum equal to 360o
.
 Suppose; for a station B.
1
2
3
4
5
6
7

4. Station Adjustment
Angles Observed Value Correction Corrected Value
∟3 --- -12'' ---
∟4 --- -12'' ---
∟5 --- -12'' ---
∟6 --- -12'' ---
∑ = 360o
00' 48'' ∑ = 360o
00' 00''

5. 2. Figure Adjustment:
 The determination of most probable values of angles involved
in any geometrical figure so as to fulfill the geometrical
conditions is called the figure adjustment. All cases of figure
adjustment necessarily involve one or more conditional
equations. The geometrical figures used in a triangulation
system are:
a. Triangles
b. Quadrilaterals
c. Polygons with central stations

6. a. Triangle Adjustment
 Triangulation of ordinary precision, the sum of the angles of a
triangle is equal to 180o
. For large triangles, covering big area,
correction for spherical excess is to be applied because sum of
angles of a spherical triangle will be more than 180o
and the
correction to be applied is as an addition for 01'' for every 75
square miles.
 For triangle ABC
Angles Observed Value Correction Corrected Value
∟1 --- -4'' ---
∟5 --- -4'' ---
∟6 --- -4'' ---
∑ = 180o
00' 12'' ∑ = 180o
00' 00''

7. b. Adjustment of Braced Quadrilateral
1. Geometric Condition:
(a) Sum of all the angles should be equal to 360o
.
(b) Sum of equal pair of angle should be equal.
∟2 + ∟3 = ∟6 + ∟7
∟1 + ∟8 = ∟4 + ∟5
 Suppose L.H.S > R.H.S by 12''
 Divide this 12” by 4; correction = 12''/4 = 3''
 Add 3'' to the angles of R.H.S and subtract 3'' from the angles
of L.H.S.

8. b. Adjustment of Braced Quadrilateral

9. 2. Trigonometric Condition:
 log (sin 1) + log (sin 3) + log (sin 5) + log (sin 7) =
log (sin 2) + log (sin 4) + log (sin 6) + log (sin 8)
 For this adjustment following procedure is adopted.
b. Adjustment of Braced Quadrilateral

10. Procedure:
 Record ‘log (sinϑ)’ of each angle obtained after geometric
adjustment.
 For each angle record the ‘log (sinϑ)’ difference for 01'' i.e.,
the difference between the previous value in step (1) & the
value obtained after adding 01'' to actual angle.
 Find the average required change (α) in ‘log (sinϑ)’ by
dividing the difference of sum of odd and even angle values
by 8.

11. Procedure
 Find the average difference “β” for difference for 01''. i.e.,
total ‘log (sin ϑ)’ difference for 01'' of all angles divided by 8.
 The ratio α / β gives the no. of seconds to apply a correction.
 Add the correction to each of four angles where ‘log (sin ϑ)’
is smaller & subtract the correction from the other four angles.

12. Problems:
 Given the observed angles of a braced quadrilateral
shown in figure. Adjust it geometrically as well as
trigonometrically.
Angle Observed Value
a 38°44'06"
b 23°44'38"
c 42°19'09"
d 44°52'01"
e 69°04'21"
f 39°37'48"
g 26°25'51"
h 75°12'14"

13. Solution:
Angle Observed Value
Figure Adjustment
Log (sin )+10ϑ
Log [sin( +01")]ϑ
+10
Log (sin )ϑ
differnece for 01"
Corrected Values
Condition 1 Condition 2
a 38°44'06"
38°44’05"
38° 44`5.5`` 9.796378 9.796380 .000002
b 23°44'38"
23° 44‘37"
23°44`35`` 9.604912 9.604917 .000005
c 42°19'09"
42° 19‘08"
42°19`06`` 9.828175 9.828178 .000003
d 44°52'01"
44° 52‘00"
44°51`59.5`` 9.848470 9.848473 .000003
e 69°04'21"
69° 04‘20"
69°04`19.5` 9.970361 9.970361 .000000
f 39°37'48"
39° 37‘47"
39°37`49`` 9.804705 9.804708 .000003
g 26°25'51"
26° 25‘50"
26°25`52`` 9.648478 9.648482 .000004
h 75°12'14"
75° 12‘13"
75°12`13.5`` 9.985355 9.985355 .000000
360°00`08`` 360°00` 39.243442 -
39.243392=.
000050//8=6
.25×10^-6
2.5*10^-6

14. Assignment:
 Given the observed angles of a braced quadrilateral
shown in figure. Adjust it geometrically as well as
trigonometrically.
Angle
Observed
Value
1 56°06'33"
2 31°56'58"
3 27°43'52"
4 63°30'12"
5 56°50'50"
6 25°23'45"
7 34°20'45"
8 64°08'54"

15. angle Observed Cond 1 Cond 2 Log
(sin )+10ϑ
Log
[sin( +01ϑ
")]+10
Corrected
Values
1 56°06'33"
2 31°56'58"
3 27°43'52"
4 63°30'12"
5 56°50'50"
6 25°23'45"
7 34°20'45"
8 64°08'54"
360°1'49"

16. SATELLITE STATION
 A satellite station is used when instrument can not
be set up at the main station. The distance of the
satellite from its station is usually very small as
compared to the length of the sides of the
triangulation.
 In the figure ABC represent part of a triangulation
station. Station A could be observed but it was not
possible to set up an instrument there. A satellite
station ‘S’ was therefore used and angles from S to
A, B & C are measured. Suppose distance AS = S'.

18. Satellite station
In the triangle ACS:
sinδ1 / S' = sinθ1 / l1
sinδ1 = S' / l1 x sin θ1----------- (1)
For very small value of δ1:
sinδ1 = δ1 (in radians)
= (δ1 x 180) / π (in degrees)
= (δ1 x 180 x 60 x 60) / π (seconds)
= 206265 x δ1 (seconds)

19. Equation (1) can be written as:
δ1 = (S' x sinθ1) / l1 (in radians)
δ1 = (206265 x S' x sinθ1) / l1 (in seconds)
Similarly:
δ2 = (206265 x S' x sinθ2) / l2 (in seconds)
BSC = θ and BAC =?
Since; α + β + θ = (α + δ1) + (β + δ2) + BAC.
Therefore;
BAC = θ - (δ1 + δ2) [For station inside the triangle]
BAC = θ + (δ1 + δ2) [For station outside the triangle]
Satellite station

20. PROBLEM
A, B & C were stations of a minor triangle ABC; C not being
suitable for an instrument. A satellite station ‘S’ was therefore
set up outside the triangle ABC in order to determine the
angle at C. The distance of ‘S’ from C was 17.00 m. The
length of AC = 16479 m and of BC = 21726 m. The angle
ASC was found to be 63o
48' 00'' and the angle ASB 71o
54'
32''. Calculate angle ACB?

21. Solution:
Given Data:
S' = 17.00 m Angle ASC = θ2 = 63o
48' 00''
Angle ASB = 71o
54' 32''
θ1 = CSB = ASB – θ2
= 71o
54' 32'' – 63o
48' 00''
θ1 = 08o
06' 32''
AC = l2 = 16479 m
BC = l1 = 21726 m
To Determine:
Angle ACB = θ =?

22. Calculations:
Using the relations:
sinδ2 = (S' x sin θ2) / l2
= [17 x sin (63o
48' 00'')] / 16479
= sin-1
[9.256 x 10-4
]
δ2 = 0o
3' 10.92''
Similarly:
Sinδ1 = (S' / l1) x sin θ2
= [17 x sin (08o
06' 32'')] / 21726
= sin-1
[1.1037 x 10-4
]
δ1 = 0o
0' 22.77''

23. Now:
Since,
Angle ACB = θ
α + β + θ = (α + δ1) + (β + δ2) + ASB
θ = δ1 + δ2 + ASB
= 0o
3' 10.92'' + 0o
0' 22.77'' + 71o
54' 32''
= 71o
58' 5.69'' (Answer)

24. Assignment
In a triangle ABC, Station C was a church spire &
could not be occupied. A satellite station was
selected at 12.10 m from ‘C’ and inside the triangle
ABC. From ‘S’ angle CSA = 135o
40' 30'' and ASB
= 71o
29' 30'' were measured. And the lengths AC
and BC were known to be approximately 2511 m
and 1894 m respectively. Compute the angle ACB?