1. Angular Measurement Requires three points Established or Determined
From – Backsight reference
At – Where the transit is located
To – Foresight
Established or Determined
Units of Measure
Degrees, Min, Sec
Radians -  rad = 180 degrees
Gons – European, 400 Gon/circle

2. Direction of a Line Same units as Angles Meridians – North/South Lines
True
Magnetic
Grid
Assumed
Azimuth
Bearing

3. Azimuth Angles Measured Clockwise from North Angle between 0 and 360°
Military and Astronomers use South
Angle between 0 and 360°
Back Az = Az ± 180°
AzBC = Back AzAB + <A

4. Bearings Quadrants – NE, SE, SW, NW Always < 90 Written N23°15’W
Back Bearing – change directions
Bearing AB = N23°15’W
Back bearing AB = bearing BA =S23°15’E

5. Bearings to Azimuths Each quadrant is different
NE: Bearing E from N Az = Bearing Angle (BA) NW NE
SE: Bearing E from S Az = 180 – BA
SW: Bearing W from S Az = BA
NW: Bearing E from N Az = 360 – BA SW SE

6. Problem 9-3, McCormac’s 4th
Az OA = 141°16’ (SE Quad)
B.A. = 180° - 141°16’= 38°84’
Bearing = S38°84’E
Az OB = 217°23’ (SW Quad)
B.A. = 217°23’ - 180° = 37°23’
Bearing = S37°23’W
Az OC = 48°23’ (NE Quad)
B.A. = Az; Bearing = N48°23’E

7. Problem 9-5, McCormac’s 4th
Az OA = 17°22’16”
Az OB = 180° - 70°18’46” = 109°41’14”
Az OB = 180° + 11°8’52” = 191°8’52”
Az OB = 360° - 76°30’52” = 283°29’8”

8. Traverse Directions Given: Find: Determine back azimuth Add angle
The direction of a side
The connecting angle
Find:
The direction of the adjacent side
Determine back azimuth
Add angle
Clockwise: (+)
Counterclockwise: (-)

9. Traverse Azimuths Example
Az AB = 35°15’
ABC = 98°27’
Find AZ BC A B C
98°27’
215°15’
Back Az = Az  180°
Az BA = 215°15’
35°15’
215°15’ + 98°27’ = 313°42’

10. Problem 9-7, McCormac’s 4th
Az AB = 70°42’ Az BA = 250°42’
ABC = 97°18’ Left (CCW)
so Az BC = 250°42’ - 97°18’ = 153°24’
SE Quadrant: B.A. = 180° - 153°24’ = 26°36’
Bearing = S 26°36’E N
250°42’ N
70°42’ A B
97°18’ C

11. Problem 9-7, McCormac’s 4th
Az BC = 162°28’ Az BA = 342°28’
BCD = 67°38’ Left (CCW)
so Az CD = 342°28’ - 67°38’ = 274°50’
NW Quadrant: B.A. = 360° - 274°50’ = 85°10’
Bearing = N 85°10’W

12. Problem 9-17, McCormac’s 5th
Az 34 = 351°50’00” Az 43 = 171°50’00” -  °37’56” °47’56” °00’00” Az 41 = 310°12’04”
NW Quad, B.A. = 360° - Az Bearing 41 = N 49°47’56”W

13. Problem 9-17, McCormac’s 5th
Az 41 = 310°12’04”
Az 14 = 130°12’04” -  °16’00” Az 12 = 78°56’04”
NE Quad, B.A. = Az Bearing 12 = N 78°56’04”E
Az 21 = 258°56’04” -  °22’00” Az 23 = 222°34’04”
SW Quad, B.A. = Az - 180° Bearing 23 = S42°34’04”E

14. Problem 9-17, McCormac’s 5th
3 = 360° - 221°37’56” °16’00” - 36°22’00” = 50°44’04”
Az 32 = 42°34’04” -  °44’04” °10’00” °00’00” Az 34 = 351°50’00”
NW Quad, B.A. = 360° - Az Bearing 34 = N 8°10’00”W
Closed traverse, Interior angles
’s = (n-2) = (4-2)180 = 360°