Surveying

2. NOTE:
To solve the first two problems, I would suggest you to first learn the
fundamentals of azimuths (also known as whole circle bearings) and bearings
(also known as quadrantal bearings). For the third problem, you need to
understand the fundamentals of mean, standard deviation and the standard
error for mean. I will try giving these fundamentals in the following paragraphs
so that you will understand how exactly I am helping you to solve the problems
given

4. The above calculation is for population standard deviation. If we
want to have sample standard deviation, we should have (n-1) in
the place of (n) in the denominator of the equation given above.

7. Hope you have gone through the pages through 3 to 6 of this file and understood the fundamentals
with which we can solve the given problems
Now
Rememberthe following N
W
S
E
I Quadrant00
TO 900
II Quadrant900
TO 1800
IIIQuadrant 1800
TO 2700
IV Quadrant 2700
TO 3600

8. 1(a) 1870
34’ 18”
As we can see here,thisazimuthliesin3rd
quadrant(SW)
3rd
Quadrantis definedbySWdirection
Bearing of this line will be S 70
34’ 18” W
N
1870
34’ 18”
1(b) 130
18’42”
As we can see here,thisazimuthliesin1st
quadrant(NE)
1st
QuadrantisdefinedbyNEdirection
So the Bearingof this line will be N 130
18’ 42” E
N
130
18’42”
1(c) 2880
44’22”
As we can see here,thisazimuthliesin4th
quadrant(NW)
4th
Quadrantis definedbyNWdirection
If you take the angle starting fromNorth andmove
towards west, this line will make
3600
00’00” - 2880
44’22” = 710
15’38” from
North direction towards West
So the Bearing of this line will be N 710
15’ 38” W
(I suggest you to draw the lines representing the
azimuth given on the quadrants given in the next
column and see for yourself the bearing given
above)
N
2880
44’22”
1(d) 1500
10’ 13”
As we can see here,thisazimuthliesin2nd
quadrant(SE)
2nd
QuadrantisdefinedbySEdirection
If you take the angle startingfromSouth andmove
towards East, this line will make
1800
00’00” - 1500
10’13” = 290
49’47” from
South direction towards East
So the Bearing of this line will be S 290
49’ 47” E
(I suggest you to draw the lines representing the
azimuth given on the quadrants given in the next
column and see for yourself the bearing given
above)
N
1500
10’ 13”

9. 2(a) N 540
34’ 37” W is in 4th
quadrant (NW)
This angle is from N towards W N
540
34’ 37”
The azimuth must be measured from North in clock wise direction always
So the azimuth will be 3600
00’ 00” - 540
34’ 37” = 3050
25’ 23”
2(b) S 090
19’ 34” W is in 3rd
quadrant (SW)
The azimuth must be measured from North in clock wise direction always
So the azimuth will be 1800
00’ 00” + 090
19’ 34” = 1890
19’ 34”
N
090
19’ 34”

10. 2(c) S 390
49’ 25” E is in 2nd
quadrant (SE)
The azimuth must be measured from North in clock wise direction always
So the azimuth will be 1800
00’ 00” - 390
49’ 25” = 1400
10’ 35”
N
390
49’ 25”
2(d) N 410
41’ 41”E is in 1st
Quadrant
410
41’ 41”
The azimuth must be measured from North in clock wise direction always
So the azimuthwill be 410
41’ 41” onlyas there isno correctionisneededforthe bearing available in 1st
quadrant.

11. As per the formulae given above, the following can be calculated
MEAN:
Mean of angle a =
(870
15’ 28” + 870
15’ 35” + 870
15’ 38” + 870
15’ 42” + 870
15’ 30”) / 5
= 870
15’ 34.6”
Mean of Angle b =
(1150
48’ 52” + 1150
48’ 58” + 1150
49’ 04” + 1150
49’ 05” + 1150
49’ 07”+ 1150
48’ 55”)/6
= 1150
49’ 0.1667”
STANDARD DEVIATION
Standard Deviation of angle a =
5.122499”
Standard Deviation of Angle b =
5.520165”
Standard Error of the Mean Values
Standard Error of the mean values is as follows for angle a =
= σ/(n)0.5
= 5.122499 / (5)0.5
= 2.290851
Standard Error of the mean values is as follows for angle b =
= σ/(n)0.5
= 5.520165 / (6)0.5
= 2.253598
All the very best