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- R0/0/639/199
- text/23/3
- Take, for instance, the following examples:
- ~
-
- L89/25/89/95
- L9/55/181/55
- L95/52/130/40
- L148/35/179/25
- L182/27/194/44
- L198/51/208/66
- text/84/16
- N
- ~
-
- text/73/45
- 1
- ~
-
- text/183/15
- 2
- ~
-
- text/199/68
- 3
- ~
-
- text/237/18
- From your starting point (1), you travel at
- a bearing of 45 for 1320' to point 2. Then
- at 165 for 1320' to point 3. What bearing
- and distance do you travel to return to 1 ?
- Note that the angle between the 2
- bearings traveled so far is 60 (if you were
- traveling straight south-180- the angle
- formed would be 45, since you are traveling
- 15 more east-165-the angle formed is
-
-
-
-
-
- ~
-
- text/14/100
- increased by 15). The 2 "legs" are of equal length. Therefore if
- you form another 60 angle between the leg from 2 to 3 and the leg
- from 3 to 1, you will be completing an equilateral triangle (3
- equal sides and 3 - 60 angles). Perhaps the easiest way to
- determine the bearing is to "look back" and determine that the
- bearing from 3 to 2 is 15 less than 360 (north), 345 . To form a 60
- angle with this line you would then travel at a bearing of 285
- (345-60) for 1320' to return to point 1. Although this explanation
- is quite involved, the actual process is relatively easy.
-
- ~
-
- C375/27/3
- C293/36/3
- C509/63/3
- C490/72/3
- C402/81/3
- C258/91/3
- C382/90/3
- C149/98/3
- C186/109/3
- C213/127/3
- C242/146/3
- C368/145/3
- C484/145/3
- C621/145/3
- C574/155/3
- C50/163/3
- C76/163/3
- C99/59/3
- C191/70/3
- C147/65/3
- C124/62/3
-
- C170/67/3
-