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

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

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

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

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 = 180 + BA NW: Bearing E from N Az = 360 – BA SW SE

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

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”

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: (-)

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’

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

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

Problem 9-17, McCormac’s 5th Az 34 = 351°50’00” Az 43 = 171°50’00” - 4 - 221°37’56” - 49°47’56” + 360°00’00” Az 41 = 310°12’04” NW Quad, B.A. = 360° - Az Bearing 41 = N 49°47’56”W

Problem 9-17, McCormac’s 5th Az 41 = 310°12’04” Az 14 = 130°12’04” - 1 - 51°16’00” Az 12 = 78°56’04” NE Quad, B.A. = Az Bearing 12 = N 78°56’04”E Az 21 = 258°56’04” - 2 - 36°22’00” Az 23 = 222°34’04” SW Quad, B.A. = Az - 180° Bearing 23 = S42°34’04”E

Problem 9-17, McCormac’s 5th 3 = 360° - 221°37’56” - 51°16’00” - 36°22’00” = 50°44’04” Az 32 = 42°34’04” - 3 - 50°44’04” - 8°10’00” + 360°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)180 = (4-2)180 = 360°